Longest Common Substring

Table of Contents

Task

Given two strings text1 and text2, return the length of their longest common subsequenceIf there is no common subsequence, return 0.

subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

  • For example, "ace" is a subsequence of "abcde".

common subsequence of two strings is a subsequence that is common to both strings.

 

Example 1:

Input:

 text1 = "abcde", text2 = "ace" 

Output:

 3  

Explanation:

 The longest common subsequence is "ace" and its length is 3.

Example 2:

Input:

 text1 = "abc", text2 = "abc"

Output:

 3

Explanation:

 The longest common subsequence is "abc" and its length is 3.

Example 3:

Input:

 text1 = "abc", text2 = "def"

Output:

 0

Explanation:

 There is no such common subsequence, so the result is 0.

 

Constraints:

  • 1 <= text1.length, text2.length <= 1000
  • text1 and text2 consist of only lowercase English characters.

This problem was taken from Leetcode Longest Common Subsequence

 

Solution

Dynamic programming with memoization.

We create a matrix to store (memoize) the response from previous calculations so we won’t re-calculate them again.

We are starting from the first row comparing every character from column 1 with the first character from row one.

There are two cases:

  • either the character match, then we add 1 and add it with the value from the diagonal to the left of the cell where we are.

Longest Common Substring Step 1

  • the characters don’t match, then we get the MAX of the value above current cell  and the value to the left of the current cell.

Longest Common Substring Step 2

Here again c int the column (first string) matches c in the row (the second string)

so we get the value of last comparison (to the left and up diagonal of the current cell)  which is 1 and add 1 again since characters match.

Longest Common Substring 3

and keep going till we reach the end of both strings which is the answer.

 

Longest Common Substring 4

 

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    var txt1 = text1.split('');
    var txt2 = text2.split('');

    txt1.unshift('0');
    txt2.unshift('1');

    var l1 = txt1.length;
    var l2 = txt2.length;

    var matrix = [];
    for(var i = 0; i < l2; i ++) {
        matrix[i] = new Array(l1).fill(0);
    }

    var maxCommon = 0;
    
    for(var row = 0; row < l2; row ++) {
        for(var col = 0; col < l1; col ++) {
            var last = 0;

            if(txt1[col] == txt2[row]) {
                var previousDiagonalRowVal = row == 0 || col == 0  ? 0 : matrix[row - 1][col - 1];
                last =  1 + previousDiagonalRowVal;
            }
            else {
                var prevUp = row == 0 ?  0 : matrix[row - 1][col];
                var prevLeft = col == 0 ? 0 : matrix[row][col - 1];               
                last = Math.max(prevUp, prevLeft);
            }
            matrix[row][col] = last;
            maxCommon = last > maxCommon ? last : maxCommon;
    
        }
    }    
    return maxCommon;
};

var text1 = "abcde", text2 = "ace";
var r = longestCommonSubsequence(text1, text2);
console.log(">>", r);

Copy List with Random Pointer

Task

 

A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

Return the head of the copied linked list.

The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

  • val: an integer representing Node.val
  • random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.

Your code will only be given the head of the original linked list.

 

Example 1:

Input:

 head = [[7,null],[13,0],[11,4],[10,2],[1,0]]

Output:

 [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

Input:

 head = [[1,1],[2,1]]

Output:

 [[1,1],[2,1]]

Example 3:

Input:

 head = [[3,null],[3,0],[3,null]]

Output:

 [[3,null],[3,0],[3,null]]

 

Constraints:

  • 0 <= n <= 1000
  • -104 <= Node.val <= 104
  • Node.random is null or is pointing to some node in the linked list.

This problem was taken from https://leetcode.com/problems/copy-list-with-random-pointer/

Solution

 

Brute force using tree traversal

 

/**
 * // Definition for a Node.
 * function Node(val, next, random) {
 *    this.val = val;
 *    this.next = next;
 *    this.random = random;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
var copyRandomList = function(head) {


    function dfsTraverse(node, visited={}) {
        if(!node) {
            return null;
        }
        
        // if a new node is created, return it. Otherwise you will fall into circular loops
        if(node?.clone) {
            return node?.clone;    
        }
        
        var newNode = new Node(node?.val, null, null);
        node.clone = newNode;
        var next = dfsTraverse(node?.next);
        var random = dfsTraverse(node?.random);

        newNode.next = next;
        newNode.random = random;
        return newNode;
    }

    var result = dfsTraverse(head);
    return result;
};

function Node(val, next, random) {
    this.val = val;
    this.next = next;
    this.random = random;
}
;// [1,null],[2,0],[3,1]

var nodes = {};
nodes['1'] = new Node(1,null,null);
nodes['2'] = new Node(2,null,null);
nodes['3'] = new Node(3,null,null);

nodes['1'].next = nodes['2'];
nodes['1'].random = null;

nodes['2'].next = nodes['3'];
nodes['2'].random = nodes['1'];

nodes['3'].next = null;
nodes['3'].random = nodes['2'];

//console.log("root");
//console.log(nodes['7']);
var result = copyRandomList(nodes['1']);
console.log(result);

 

 

More elegant solution

 

/**
 * // Definition for a Node.
 * function Node(val, next, random) {
 *    this.val = val;
 *    this.next = next;
 *    this.random = random;
 * };
 */

/**
 * @param {Node} head
 * @return {Node}
 */
var copyRandomList = function(head) {
  let cur = head;
  const copy = new Map();

  // add all new nodes and values for now
  while (cur) {
    copy.set(cur, new Node(cur.val));
    cur = cur.next;
  }
  
  cur = head;

  // iterate again and point curent node to the newly created nodes using the key 
  while (cur) {
    copy.get(cur).next = copy.get(cur.next) || null;
    copy.get(cur).random = copy.get(cur.random) || null;
    cur = cur.next;
  }
  
  return copy.get(head);
}


function Node(val, next, random) {
    this.val = val;
    this.next = next;
    this.random = random;
};


// [1,null],[2,0],[3,1]

var nodes = {};
nodes['1'] = new Node(1,null,null);
nodes['2'] = new Node(2,null,null);
nodes['3'] = new Node(3,null,null);

nodes['1'].next = nodes['2'];
nodes['1'].random = null;

nodes['2'].next = nodes['3'];
nodes['2'].random = nodes['1'];

nodes['3'].next = null;
nodes['3'].random = nodes['2'];


var result = copyRandomList(nodes['1']);
console.log(result);

 

 

Range Sum of BST

Task

Given the root node of a binary search tree and two integers low and high, return the sum of values of all nodes with a value in the inclusive range [low, high].

 

Example 1:

Input:

 root = [10,5,15,3,7,null,18], low = 7, high = 15

Output:

 32

Explanation:

 Nodes 7, 10, and 15 are in the range [7, 15]. 7 + 10 + 15 = 32.

Example 2:

Input:

 root = [10,5,15,3,7,13,18,1,null,6], low = 6, high = 10

Output:

 23

Explanation:

 Nodes 6, 7, and 10 are in the range [6, 10]. 6 + 7 + 10 = 23.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 2 * 104].
  • 1 <= Node.val <= 105
  • 1 <= low <= high <= 105
  • All Node.val are unique.

This problem was taken from https://leetcode.com/problems/range-sum-of-bst/

 

Solution

 

DFS Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} low
 * @param {number} high
 * @return {number}
 */
var rangeSumBST = function(root, low, high) {
    if(!root) {
        return;
    }
    console.log(root.val);
    var queue = [root];
    var visited = {};
    var result = 0.0;    

    function dfsTraverse(root, low, high) {
    
        var childNodes = [root?.left, root?.right];

        for(const childNode of childNodes) {
          if(childNode) {
            var val = childNode.val;
            console.log(val);
            dfsTraverse(childNode, low, high);            
          }
        }

    }

    dfsTraverse(root, low, high);
    return result;
};


 function TreeNode(val, left, right) {
     this.val = (val===undefined ? 0 : val)
     this.left = (left===undefined ? null : left)
     this.right = (right===undefined ? null : right)
 }

const treeNode = new TreeNode(
    10,
    new TreeNode(
        5,
        new TreeNode(3),
        new TreeNode(7)
    ),
    new TreeNode(
        15,
        null,
        new TreeNode(18)
    )
)


rangeSumBST(treeNode, 7, 15);

 

BFS solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} low
 * @param {number} high
 * @return {number}
 */
var rangeSumBST = function(root, low, high) {
    if(!root) {
        return 0;
    }
    var queue = [root];
    var visited = {};
    var result = 0.0;    
    if(root.val >= low && root.val <= high) {
        result = root.val;
    }
    
    var leftPointer = 0;
    while(leftPointer <= queue.length) {
        var curentNode = queue[leftPointer];
        leftPointer ++;
        var childNodes = [curentNode?.left, curentNode?.right];
        
        for(const childNode of childNodes) {
            if(childNode) {
                console.log(childNode.val)
                if(childNode.val >= low && childNode.val <= high) {
                    result += childNode.val;
                }
                queue.push(childNode);
            }
        }
        
           
    }
    return result;
    
};

 

Graph traversal

The graph

The graph represent connections between airports. Nodes (vertices)  represent airports, and  edges represent flights.

BFS search (Breadth First Search)

Breadth-first Search (BFS) starts by pushing all of the direct children to a queue (first-in, first-out). It then visits each item in queue and adds the next layer of children to the back of the queue. Since one node could be visited multiple times causing infinite loop, we have to keep track of all visited nodes.

Using JavaScript Map

// DATA
const airports = 'PHX BKK OKC JFK LAX MEX EZE HEL LOS LAP LIM'.split(' ');

const routes = [
    ['PHX', 'LAX'],
    ['PHX', 'JFK'],
    ['JFK', 'OKC'],
    ['JFK', 'HEL'],
    ['JFK', 'LOS'],
    ['MEX', 'LAX'],
    ['MEX', 'BKK'],
    ['MEX', 'LIM'],
    ['MEX', 'EZE'],
    ['LIM', 'BKK'],
];


// The graph
const adjacencyList = new Map();

// Add node
function addNode(airport) {
    adjacencyList.set(airport, []);
}

// Add edge, undirected
function addEdge(origin, destination) {
    adjacencyList.get(origin).push(destination);
    adjacencyList.get(destination).push(origin);
}

// Create the Graph
airports.forEach(addNode);
routes.forEach(route => addEdge(route[0], route[1]));

console.log(adjacencyList);


function bfs(start, searchItem) {

    const visited = new Set();
    const queue = [start];


    while (queue.length > 0) {

        const airport = queue.shift(); // mutates the queue
        const destinations = adjacencyList.get(airport);


        for (const destination of destinations) {
            if (destination === searchItem)  {
                console.log(`BFS found a route to`, searchItem)
            }

            if (!visited.has(destination)) {
                visited.add(destination);
                queue.push(destination);
            }           
        }        
    }

}

bfs('PHX', 'BKK');

 

Using JavaScript object

// Data
var airports = {
    'PHX': ['LAX', 'JFK'],
    'BKK': ['MEX', 'LIM'],
    'OKC': ['JFK'],
    'JFK': ['LAX', 'PHX', 'OKC', 'HEL', 'LOS', 'EZE'],
    'LAX': ['PHX', 'JFK', 'MEX'],
    'MEX': ['LAX', 'EZE', 'BKK', 'LIM'],
    'EZE': ['JFK', 'MEX'],
    'HEL': ['JFK'],
    'LOS': ['JFK'],
    'LAP': [],
    'LIM': ['MEX', 'BKK']
};


function bfs(start, endDest) {
    var queue = [start];
    var visited = {};

    while(queue.length > 0) {
        var curentNodeVal = queue.shift();
        var childNodes = airports[curentNodeVal];
        for(const childNode of childNodes) {
            if(childNode === endDest) {
                console.log("BFS found a route to :", endDest);
            }            

            if(!visited[childNode]) {
                console.log(childNode);
                visited[childNode] = true;
                queue.push( childNode);          
            }
        }
    }
}


bfs('PHX', 'BKK');

 

DFS (Depth First Search)

Depth-first Search (DFS) will go as far into the graph as possible until it reaches a node without any children, at which point it backtracks and continues the process. The algorithm can be implemented with a recursive function that keeps track of previously visited nodes. If a node has not been visited, we call the function recursively.

Using JavaScript Map

// DATA
const airports = 'PHX BKK OKC JFK LAX MEX EZE HEL LOS LAP LIM'.split(' ');

const routes = [
    ['PHX', 'LAX'],
    ['PHX', 'JFK'],
    ['JFK', 'OKC'],
    ['JFK', 'HEL'],
    ['JFK', 'LOS'],
    ['MEX', 'LAX'],
    ['MEX', 'BKK'],
    ['MEX', 'LIM'],
    ['MEX', 'EZE'],
    ['LIM', 'BKK'],
];

// The graph
const adjacencyList = new Map();

// Add node
function addNode(airport) {
    adjacencyList.set(airport, []);
}

// Add edge, undirected
function addEdge(origin, destination) {
    adjacencyList.get(origin).push(destination);
    adjacencyList.get(destination).push(origin);
}

// Create the Graph
airports.forEach(addNode);
routes.forEach(route => addEdge(route[0], route[1]));

console.log(adjacencyList);



function dfs(start, visited = new Set()) {

    console.log(start)
    
    visited.add(start);

    const destinations = adjacencyList.get(start);

    for (const destination of destinations) {

        if (destination === 'BKK') { 
            console.log(`DFS found Bangkok`)
            return;
        }
        
        if (!visited.has(destination)) {
            dfs(destination, visited);
        }

    }

}

dfs('PHX');

Using JavaScript object

// Data
var airports = {
    'PHX': ['LAX', 'JFK'],
    'BKK': ['MEX', 'LIM'],
    'OKC': ['JFK'],
    'JFK': ['LAX', 'PHX', 'OKC', 'HEL', 'LOS', 'EZE'],
    'LAX': ['PHX', 'JFK', 'MEX'],
    'MEX': ['LAX', 'EZE', 'BKK', 'LIM'],
    'EZE': ['JFK', 'MEX'],
    'HEL': ['JFK'],
    'LOS': ['JFK'],
    'LAP': [],
    'LIM': ['MEX', 'BKK']
};


function dfs(start, endDest, visited = {}) {
        
    visited[start] = true;
    console.log(start);

    const destinations = airports[start];

    for(const destination of  destinations) {
        if (destination === endDest) { 
            console.log(`DFS found route to`, endDest);
        }

        if(!visited[destination]) {
            dfs(destination, endDest, visited);
        }
        
    }
}


dfs('PHX', 'BKK');

 

Moving Average from Data Stream

Table of Contents

Task

Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.

Implement the MovingAverage class:

  • MovingAverage(int size) Initializes the object with the size of the window size.
  • double next(int val) Returns the moving average of the last size values of the stream.

 

Example 1:

Input

["MovingAverage", "next", "next", "next", "next"]
[[3], [1], [10], [3], [5]]

Output

[null, 1.0, 5.5, 4.66667, 6.0]

Explanation

MovingAverage movingAverage = new MovingAverage(3);
movingAverage.next(1); // return 1.0 = 1 / 1
movingAverage.next(10); // return 5.5 = (1 + 10) / 2
movingAverage.next(3); // return 4.66667 = (1 + 10 + 3) / 3
movingAverage.next(5); // return 6.0 = (10 + 3 + 5) / 3

 

Constraints:

  • 1 <= size <= 1000
  • -105 <= val <= 105
  • At most 104 calls will be made to next.

Solution

/**
 * @param {number} size
 */
var MovingAverage = function(size) {
    this.n = size;
    this.queue = [];
    this.average = 0.0;
};

/** 
 * @param {number} val
 * @return {number}
 */
MovingAverage.prototype.next = function(val) {
    var removedVal;
    if(this.queue.length >= this.n) {
        removedVal = this.queue.shift();
        this.average = this.average - removedVal;
    }

    this.queue.push(val);
    
    this.average += val;
    console.log(this.average / this.queue.length);
    return this.average / this.queue.length;
};


var movingAverage = new MovingAverage(3);
movingAverage.next(1); // return 1.0 = 1 / 1
movingAverage.next(10); // return 5.5 = (1 + 10) / 2
movingAverage.next(3); // return 4.66667 = (1 + 10 + 3) / 3
movingAverage.next(5); // return 6.0 = (10 + 3 + 5) / 3

 

Heaps or priority queues in Java Script

Still work in progress …

 

Part I Heapify …

 

heap-sorting

 

Part II sort and heapify

We start swapping values and heapify each time starting from 0 to the length of i

function heapSort(arr) {

    function maxHeapify(arr, i, N) {
        var leftChild = i * 2 + 1;
        var rightChild = i * 2 + 2;
        var largest = i;
    
        if(leftChild < N && arr[leftChild] > arr[largest]) {
            largest = leftChild;
        }
    
        if(rightChild < N && arr[rightChild] > arr[largest]) {
            largest = rightChild;
        }
    
        if(largest != i) {
            var temp = arr[i];
            arr[i] = arr[largest];
            arr[largest] = temp;
            maxHeapify(arr, largest, N);
        }
    }
    
    // CREATE A HEAP
    for(var i = parseInt(arr.length / 2 - 1); i >= 0; i--) {
        maxHeapify(arr, i, arr.length);
    }
    
    console.log("heap represented in array: ", arr);
    
    // SORT THE ARRAY
    for(var i =  arr.length - 1; i >= 0; i --) {
        var temp = arr[0];
        arr[0] = arr[i];
        arr[i] = temp;
    
        maxHeapify(arr, 0, i);
    }
}

const arr = [4, 6, 3, 2, 9, 1];
heapSort(arr);

console.log("sorted array:", arr);

 

Implement pow(x, n)

Task

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

 

Example 1:

Input:

 x = 2.00000, n = 10

Output:

 1024.00000

Example 2:

Input:

 x = 2.10000, n = 3

Output:

 9.26100

Example 3:

Input:

 x = 2.00000, n = -2

Output:

 0.25000

Explanation:

 2

-2

 = 1/2

2

 = 1/4 = 0.25

 

Constraints:

  • -100.0 < x < 100.0
  • -231 <= n <= 231-1
  • -104 <= xn <= 104

This problem was taken from https://leetcode.com/problems/powx-n/

 

Solution

 

Brute force solution:

Straight forward, but we have to consider negative ranges.

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    
    if(n === 0) {
        return 1;
    }

    var N = n;
    var X = x;
    var result = 1;
    
    if(n < 0) {
        X = 1 / x;
        N = -n;
    }
    
    for(var i = 0; i < N; i ++) {
        result = result * X;
    }  
    return result;
};

 

Approach 2: Fast Power Algorithm Recursive

  • divide n so you immediately cut the computation time in half to logarithmic one.

pow of 2 to the power of 6

 

 

/**
 * @param {number} x
 * @param {number} n
 * @return {number}
 */
var myPow = function(x, n) {
    
    function fastPow(x, n) {
        if(n < 1) {
            return 1;
        }
        var half = fastPow(x, Math.floor(n / 2));

        if(n % 2 == 0) {
            return half * half;
        }
        else {
            return half * half * x;
        }
    }
    
    var X = x;
    var N = n;
    if(n < 0) {
        X = 1 / x;
        N = -n;        
    }
    return fastPow(X, N);
    
};

 

 

regular JavaScript functions vs array functions

Does not have its own bindings to this or super, and should not be used as methods.

var obj1 = {
    a: 123,
    b: function() {
        return this.a;
    }
}

var obj2 = {
    a: 123,
    b: () => {
        return this.a;
    }
}

console.log("obj1: ", obj1.b());
console.log("obj2: ", obj2.b());

==========================================
result
==========================================

obj1:  123
obj2:  undefined

Using call/apply and bind

var obj1 = {
    a: 111,
    b: (a,b) => {
        return `${this.a} ${a} ${b}`;
    },
    c: function(a,b) {
        return `${this.a} ${a} ${b}`;        
    }
}

var obj2 = {
    a: 123
}

console.log("obj1: ", obj1.b.apply(obj2, ['one', 'two']));
console.log("obj1: ", obj1.b.bind(obj2)());


console.log("obj1: ", obj1.c.apply(obj2, ['one', 'two']));
console.log("obj1: ", obj1.c.bind(obj2, 'one', 'two')());

=====================================================
result
=====================================================
obj1:  undefined one two
obj1:  undefined undefined undefined
obj1:  123 one two
obj1:  123 one two

 

JavaScript call, bind and apply simple

According MDN all three methods are very similar, and at the end they produce very similar result.

  • they all accept this object as first parameter.
  • then apply accepts and array of arguments vs bind and call accept a sequence of arguments.

Function.prototype.apply()

The apply() method calls a function with a given this value, and arguments provided as an array

Function.prototype.bind()

The bind() method creates a new function that, when called, has its this keyword set to the provided value, with a given sequence of arguments preceding any provided when the new function is called

 

const module = {
  getX: function(one, two) {
    return `${this.x} ${one} ${two}`;
  }
};


const obj = {x: '123'};


console.log( "call: ",module.getX.call(obj, 'ONE', 'TWO') );

console.log( "apply: ",module.getX.apply(obj, ['ONE', 'TWO']) );

var newFunc = module.getX.bind( obj, 'ONE', 'TWO');
console.log("bind: ", newFunc());


result should be the same.

call:  123 ONE TWO
apply:  123 ONE TWO
bind:  123 ONE TWO

 

Longest Common Prefix

Table of Contents

Task

 

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

 

Example 1:

Input:

 strs = ["flower","flow","flight"]

Output:

 "fl"

Example 2:

Input:

 strs = ["dog","racecar","car"]

Output:

 ""

Explanation:

 There is no common prefix among the input strings.

 

Constraints:

  • 1 <= strs.length <= 200
  • 0 <= strs[i].length <= 200
  • strs[i] consists of only lower-case English letters.

This problem was taken from Leetcode Longest Common Prefix

 

Solution

 

Let’s examine this example:

 strs = ["flower","flow","flight"]

the longest common substring is:

 "fl"

Solution 1: Horizontal scanning

We could assume that the whole word could be the common one so we set prefix = ‘flower’
Then we would compare with the rest words and keep removing last character until prefix becomes empty (meaning no common substring was found) or until we have the common substring.

prefix flower flow flight
flower flower flower flower
flowe flowe flowe flowe
flow flow flow flow
flo flo flo flo
fl fl fl fl

 

/**
 * @param {string[]} strs
 * @return {string}
 */
var longestCommonPrefix = function(strs) {
    var prefix = strs[0];
    for(var i = 1; i < strs.length; i ++ ) {
        while(strs[i].indexOf(prefix) != 0) {
            prefix = prefix.substring(0, prefix.length - 1);
            if(prefix == "")
                return '';
        }
    }
    return prefix;    
};

 what we just did:
– set up prefix to be the whole 1st word strs[0]
– compare prefix with the second word (strs[1]) and if there is no match, remove the last symbol and keep going until it finds match.

Complexity Analysis

  • Time complexity : O(S) , where S is the sum of all characters in all strings.In the worst case all n strings are the same. The algorithm compares the string S1 with the other strings [S_2 \ldots S_n] There are S character comparisons, where S is the sum of all characters in the input array.
  • Space complexity : O(1). We only used constant extra space.

 

Solution 2: Vertical scanning

Similar but optimized for cases like the one above where we have very short common substring, and we don’t want to scan the whole word.

 

prefix flower flow flight
f f f f
fl fl fl fl
flo flo flo flo

 

/**
 * @param {string[]} strs
 * @return {string}
 */
var longestCommonPrefix = function(strs) {
    var prefix;    
    for(var i = 0; i < strs[0].length; i ++ ) {
        var c = strs[0][i];
        for(var j = 0; j < strs.length; j++) {
            if(strs[j][i] != c) {
                return strs[0].substring(0, i);
            }
        }
    }
    return strs[0];    
};

 what we just did:
– Iterate through the words like they are in column.
– compare each character (starting with the first one) between all words. When we find a mismatch, remove the last (mismatched) character and return truncates strs[0]