LeetCode Problem Solving Series: Problem 1 TWO SUM #leetcode

Leet Code 1. Two Sum : Hey there, coding enthusiasts! Welcome back to another exciting coding session. Today’s problem is a treat—literally! We’re going to solve the “two sum ” or “Leet Code .1‘

Time Complexity:

• Brute-force: O(n^2)
• HashMap: O(n)
• Two Pass HashMap: O(n)
• Two Pointer: O(n log n)

JavaScript:

1. Brute Force Approach (JavaScript):

function twoSum(nums, target) {
    for (let i = 0; i < nums.length; i++) {
        for (let j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] === target) {
                return [i, j];
            }
        }
    }
    return []; // No solution found!
}

2. HashMap Approach (JavaScript):

function twoSum(nums, target) {
    const numToIndex = new Map();
    for (let i = 0; i < nums.length; i++) {
        const complement = target - nums[i];
        if (numToIndex.has(complement)) {
            return [numToIndex.get(complement), i];
        }
        numToIndex.set(nums[i], i);
    }
    return []; // No solution found!
}

3. Two Pass HashMap Approach (JavaScript):

var twoSum = function(nums, target) {
    const numToIndex = {};
    
    // First pass: Populate the object
    for (let i = 0; i < nums.length; i++) {
        numToIndex[nums[i]] = i;
    }
    
    // Second pass: Check for the complement
    for (let i = 0; i < nums.length; i++) {
        const complement = target - nums[i];
        if (complement in numToIndex && numToIndex[complement] !== i) {
            return [i, numToIndex[complement]];
        }
    }
    
    return [];
};

4. Two-Pointer Approach (JavaScript):

function twoSum(nums, target) {
    const numsWithIndex = nums.map((num, index) => [num, index]);
    numsWithIndex.sort((a, b) => a[0] - b[0]);
    let left = 0, right = nums.length - 1;
    while (left < right) {
        const sum = numsWithIndex[left][0] + numsWithIndex[right][0];
        if (sum === target) {
            return [numsWithIndex[left][1], numsWithIndex[right][1]];
        } else if (sum < target) {
            left++;
        } else {
            right--;
        }
    }
    return []; // No solution found!
}

And there you have it! Four different programming languages, four approaches to solving the “Two Sum” problem, and hopefully a smile on your face as well! Remember, choosing the right approach depends on the specific requirements and constraints of your problem. Happy coding!