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

A 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".

A 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.

Idea:

Dynamic Programing

Use dp[i][j] to represent the length of longest common sub-sequence of text1[0:i-1] and text2[0:j-1]
dp[i][j] = dp[i – 1][j – 1] + 1 if text1[i – 1] == text2[j – 1] else max(dp[i][j – 1], dp[i – 1][j])

Solution:

/**
 * @param {string} text1
 * @param {string} text2
 * @return {number}
 */
var longestCommonSubsequence = function(text1, text2) {
    // Use dp[i][j] to represent the length of longest common sub-sequence of text1[0:i-1] and text2[0:j-1]
    // dp[i][j] = dp[i – 1][j – 1] + 1 if text1[i – 1] == text2[j – 1] else max(dp[i][j – 1], dp[i – 1][j])
    let dp = Array.from(new Array(text1.length + 1), () => new Array(text2.length + 1));
    
    // final result dp[i + 1][j + 1] contains length of LCS  
    // for text1[0..i] and text2[0..j], so use <=
    for (let i = 0; i <= text1.length; i++) {
        for (let j = 0; j <= text2.length; j++) {
            // base case:
            if (i === 0 || j === 0) 
                dp[i][j] = 0;
            else if (text1[i - 1] === text2[j - 1])
                dp[i][j] = dp[i - 1][j - 1] + 1;
            else
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
        }
    }
    // dp[text1.length][text2.length] contains length of LCS  
    // for text1[0..text1.length-1] and text2[0..text2.length-1]
    return dp[text1.length][text2.length];
}

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

Example 1:

Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Example 2:

Input: nums = [0,1,0,3,2,3]
Output: 4

Example 3:

Input: nums = [7,7,7,7,7,7,7]
Output: 1

Constraints:

• 1 <= nums.length <= 2500
• -104 <= nums[i] <= 104

Idea:

dp[i] represents the length of the longest increasing subsequence of nums[0...i]
dp[0] = 1
dp[i] = max(dp[j]'s) + 1 if j < i when nums[i] > nums[j]
result: find the max of all dp[i]'s

Solution:

/**
 * @param {number[]} nums
 * @return {number}
 */
var lengthOfLIS = function(nums) {
    let dp = new Array(nums.length);
    dp[0] = 1;
    for (let i = 0; i < nums.length; i++) {
        // local max
        let max = 0;
        for (let j = 0; j < i; j++) {
            if (nums[i] > nums[j]) {
                max = Math.max(dp[j], max);
            }
        }
        dp[i] = max + 1;
    }
    // result: find the max of all dp[i]'s
    return Math.max(...dp);
};

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]
Output: 7
Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]
Output: 12

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• 0 <= grid[i][j] <= 100

Idea:

Dynamic Programing

save space: we just use the original grid to calcuate the min path sum
grid[i][j] represent the min path sum of i x j grid
grid[i][j] = get the min of previous grid(top/left) + current val

Solution:

/**
 * @param {number[][]} grid
 * @return {number}
 */
var minPathSum = function(grid) {
    // save space: we just use the original grid to calcuate the min path sum
    // grid[i][j] = min path sum of i x j grid
    let m = grid.length;
    let n = grid[0].length;
    
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            // grid[0][0] itself contains the min path sum
            if (i === 0 && j === 0) continue;
            // first row: grid[i][j] = previous grid(left) + current val
            if (i === 0) grid[i][j] += grid[i][j - 1];
            // first column: grid[i][j] = previous grid(top) + current val
            else if (j === 0) grid[i][j] += grid[i - 1][j];
            // grid[i][j] = get the min of previous grid(top/left) + current val
            else grid[i][j] += Math.min(grid[i][j - 1], grid[i - 1][j]);
        }
    }
    return grid[m - 1][n - 1];
}

A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).

How many possible unique paths are there?

Example 1:

Input: m = 3, n = 7
Output: 28

Example 2:

Input: m = 3, n = 2
Output: 3
Explanation:
From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down

Example 3:

Input: m = 7, n = 3
Output: 28

Example 4:

Input: m = 3, n = 3
Output: 6

Constraints:

• 1 <= m, n <= 100
• It’s guaranteed that the answer will be less than or equal to 2 * 109.

Idea:

Dynamic Programing

dp[m][n] = unique paths of m x n grid
// base case:
dp[1][1] = 1; 
// dp[i][j] = top(unique paths) + left(unique paths)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];

Solution:

/**
 * @param {number} m
 * @param {number} n
 * @return {number}
 */
var uniquePaths = function(m, n) {
    // one row or one column dp[i][1] = dp[1][j] = 1; Let's fill them all 1
    let dp = Array.from(new Array(m), () => new Array(n).fill(1));
    for (let i = 0; i < m; i++) {
        for (let j = 0; j < n; j++) {
            // dp[i][j] = top(unique paths) + left(unique paths)
            dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
        }
    }
    return dp[m - 1][n - 1];
}

Given a string s, return the longest palindromic substring in s.

Example 1:

Input: s = "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: s = "cbbd"
Output: "bb"

Example 3:

Input: s = "a"
Output: "a"

Example 4:

Input: s = "ac"
Output: "a"

Constraints:

• 1 <= s.length <= 1000
• s consist of only digits and English letters (lower-case and/or upper-case),

Idea:

Dynamic Programing

dp[i][j] = if the substring from i to j is a palindrome = true
dp[i][j] = dp[i+1][j-1] and s[i] === s[j]
base case:
dp[i][i] = true; 
dp[i][i+1] = s[i] === s[i+1];

• Time complexity : O(n²)
• Space complexity : O(n²)

Solution:

var longestPalindrome = function(s) {
    let len = s.length;
    if (s === null || len === 1) return s;
    let dp = Array.from(new Array(len), () => new Array(len));
    let res = "";
    let max = 0;
    
    for (let j = 0; j < len; j++) {
        // remember i <= j
        for (let i = 0; i <= j; i++) {
            let isSame = s[i] === s[j];
            
            // one and two letters palindromes only check s[i] === s[j]
            // more than 2, check subset dp[][] && s[i] === s[j]
            dp[i][j] = j - i <= 2 ? isSame : dp[i + 1][j - 1] && isSame;
            
            // any new max palindrome, update max and longest result
            if (dp[i][j] && j - i + 1 > max) {
                max = j - i + 1;
                res = s.substring(i, j + 1);
            }
        }
    }
    return res;
}