CODEWITHSUNDEEP
javascriptarraysnested-loops
Jerry Osorio
Jerry Osorio

Reputation: 25

how to improve performance of nested for loops when comparing values in an array

var twoSum = function(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 [nums.indexOf(nums[i]), nums.lastIndexOf(nums[j])]
      }
    }
  }
}
function([3,5,3,7,2], 9) //It should return [3, 4] since 7 + 2 = 9

This is a Javascript function and it's iterating over the array of numbers with nested for loops and find a pair that when summed, it equals the target and return their indexes. Since it has a nested for loop it I believe it has O(n^2) or quadratic time complexity, and I was wondering if there is a faster way to do this. I'm just trying to get the first pass solution and after improve upon it. 😁 Thanks in advance!

Upvotes: 2

Views: 1407

Answers (2)

Barmar
Barmar

Reputation: 781004

Create an object that maps each element to its last index. Then loop over each element, and see if target - element is in the object with a different index.

function twoSums(nums, target) {
  const last_index = {};
  nums.forEach((num, i) => last_index[num] = i);
  for (let i = 0; i < nums.length; i++) {
    const diff = target - nums[i];
    if (diff in last_index && last_index[diff] != i) {
      return [i, last_index[diff]];
    }
  }
}

const list = [3,5,3,7,2];
console.log(twoSums(list, 9));
console.log(twoSums(list, 6));
console.log(twoSums(list, 15));

Looking up an object property is O(1), so this algorithm is O(n).

Upvotes: 5

Mosia Thabo
Mosia Thabo

Reputation: 4267

You could use nested map():

var twoSum = (nums, target)=> {
  //  map trough list for first number
  nums.map(
  	(val1, index1)=>
    {
      // map through list again for second number to compare to number returned from first map at each instance
      nums.map((val2, index2) =>
      {
      //  Ensure that index1 and index2 are not the same before you compare
        if(index1 != index2 && (val1 + val2 === target))
        {
          alert([index1, index2]);
          return [index1, index2];
        }
      });
    }
  );
}

twoSum([3,5,3,7,2], 9) //It should return [3, 4] since 7 + 2 = 9

Note: Remember this won't loop through all the items unless a match has not been found. As soon as a match is found, the function returns, and exit.

Upvotes: 2

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