JS - Educative-Twopointers-02

QUESTION: Sum of Three Values

Given an array of integers, nums, and an integer value, target, determine if there are any three integers in nums whose sum is equal to the target
Return true if three such integers exist in the array, otherwise return false.
The triplet are elements with disttinct indexes.

Explanation:

nums[i] + nums[j] + nums[k] == target;

SOLUTION:

Using 2-Pointers

const findSumOfThree = (s) => {
  // Sort the input array
  nums.sort((a, b) => a - b);
 
  const result = [];
 
  // Iterate through the array
  for (let i = 0; i < nums.length - 2; i++) {
    // Initialize two pointers
    let j = i + 1;
    let k = nums.length - 1;
 
    // While j is less than k
    while (j < k) {
      // Compute the sum of the three elements
      const sum = nums[i] + nums[j] + nums[k];
 
      if (sum === target) {
        // Add these three elements to the result array
        result.push([nums[i], nums[j], nums[k]]);
 
        // Move both pointers towards the center
        j++;
        k--;
      } else if (sum < target) {
        // Increment j if the sum is less than the target sum
        j++;
      } else {
        // Decrement k if the sum is greater than the target sum
        k--;
      }
    }
  }
 
  return result;
};
 
const nums = [-1, 0, 1, 2, -1, -4];
const target = 0;
const result = threeSum(nums, target);
console.log(result); // [ [ -1, -1, 2 ], [ -1, 0, 1 ], [ -1, 0, 1 ] ]

Time complexity

The time complexity is O(n) where n is the number of characters present in the string.

Space complexity

The space complexity is O(1) because we use constant space to store two indices.