Improving Efficiency with Nested Loop Restructuring

Question

Improving Efficiency with Nested Loop Restructuring

Directions

If you have an array of integers, the task is to find the indices of two numbers that add up to a specific target.

There will be exactly one solution for each input, and you cannot use the same element twice.

For Example

Given nums = [2, 7, 11, 15], target = 9,

Since nums[0] + nums[1] = 2 + 7 = 9, you should return [0, 1].

Is there a way to achieve this without using nested for-loops? I want to lower the time complexity.

Sample Code

const twoSum = function(nums, target) {
    for(let i in nums){
      for(let j in nums) {
        if(nums[i] + nums[j] === target && nums[i] != nums[j]) {
            return [i, j];
        }
      }
    }
};

console.log(twoSum([2, 7, 11, 15], 9));

javascript arrays for-loop nested-loops

Answer 1

Answer №1

One approach is to store the difference between each element and the target value in an object, with the differences as keys and their corresponding indices as values. This allows for quick lookup without iterating through the entire array. Then, in a separate loop, check if the elements exist in the object - if they do, you have found a pair. To avoid comparing an element with itself, an additional condition is included.

const twoSum = function(nums, target) {  
  const temp = {};
  for(let i=0; i<nums.length; i++) {
    temp[target - nums[i]] = i;
  }

  for(let i=0; i<nums.length-1; i++) {
    if(temp[nums[i]] && temp[nums[i]] !== i) {
      return [i, temp[nums[i]]]
    }
  }
};

console.log(twoSum([2, 11, 7, 17], 9));
console.log(twoSum([1, 3, 4, 2], 6));

Answer 2

One approach is to store the difference between each element and the target value in an object, with the differences as keys and their corresponding indices as values. This allows for quick lookup without iterating through the entire array. Then, in a separate loop, check if the elements exist in the object - if they do, you have found a pair. To avoid comparing an element with itself, an additional condition is included.

const twoSum = function(nums, target) {  
  const temp = {};
  for(let i=0; i<nums.length; i++) {
    temp[target - nums[i]] = i;
  }

  for(let i=0; i<nums.length-1; i++) {
    if(temp[nums[i]] && temp[nums[i]] !== i) {
      return [i, temp[nums[i]]]
    }
  }
};

console.log(twoSum([2, 11, 7, 17], 9));
console.log(twoSum([1, 3, 4, 2], 6));

Answer 3

Answer №2

Given the nature of this assignment, I will offer some guidance while refraining from revealing a complete resolution:

Your existing code involves redundant index checks. Instead of looping over [0,1] and [1,0], where the sums will always be equal since a+b = b+a, consider having your loop for i range from 0 to len-1, and your loop for j range from i+1 to len-1 to avoid duplicates.
In your current condition check nums[i] != nums[j], note that it's not specified in the problem that array values can't be identical. Is it conceivable to use values like toSum([1, 4, 4], 8) with 4+4=8? If so, eliminating the nums[i] != nums[j] check could optimize processing time.
If the provided array isn't sorted, you may want to introduce a tracking mechanism to keep record of already assessed values and skip reevaluation on subsequent iterations. For instance, if 4 has been compared against all other values without success, encountering 4 later doesn't necessitate reexamination.

Answer 4

Given the nature of this assignment, I will offer some guidance while refraining from revealing a complete resolution:

Your existing code involves redundant index checks. Instead of looping over [0,1] and [1,0], where the sums will always be equal since a+b = b+a, consider having your loop for i range from 0 to len-1, and your loop for j range from i+1 to len-1 to avoid duplicates.
In your current condition check nums[i] != nums[j], note that it's not specified in the problem that array values can't be identical. Is it conceivable to use values like toSum([1, 4, 4], 8) with 4+4=8? If so, eliminating the nums[i] != nums[j] check could optimize processing time.
If the provided array isn't sorted, you may want to introduce a tracking mechanism to keep record of already assessed values and skip reevaluation on subsequent iterations. For instance, if 4 has been compared against all other values without success, encountering 4 later doesn't necessitate reexamination.

Answer 5

Answer №3

To efficiently tackle this issue, you can utilize an algorithm with a time complexity of O(n). The prerequisite for executing this approach is that the array needs to be in sorted order.

let twosum = (arr, x) => {
  let s = 0,
    e = arr.length - 1;
  let loc = [];

  while (s < e) {
    if (arr[s] + arr[e] === x) {
      loc.push([s,e]);
      s++;
      e--;
    } else if (arr[s] + arr[e] < x) {
      s++;
    } else {
      e--;
    }
  }

  return loc;
};

console.log(twosum([1, 2, 3, 4, 5, 7, 8], 9));
console.log(twosum([2, 7, 11, 15], 9));

The step-by-step breakdown of this algorithm:

1.   Set the initial value of 's' to 0.
2.   Set the final value of 'e' to the last index (arr.length - 1).
3.   While 's' is less than 'e' meaning they have not crossed each other yet:
4.   If arr[s] + arr[e] equals x, we have found a match.
4.1. Increment 's' by 1 as there are no possible combinations before the current 's'.
4.2. Decrement 'e' by 1 as there are no possible combinations after the current 'e'.
4.3. Save the indexes where the match was located.
5.   If arr[s] + arr[e] is less than x:
5.1. Increment 's' since there are no possible combinations before the current 's', but the possibility of a match still exists for 'e'.
6.   If arr[s] + arr[e] is greater than x:
6.1. Decrement 'e' as there are no possible combinations after the current 'e', but the possibility of a match still exists for 's'.

Answer 6

To efficiently tackle this issue, you can utilize an algorithm with a time complexity of O(n). The prerequisite for executing this approach is that the array needs to be in sorted order.

let twosum = (arr, x) => {
  let s = 0,
    e = arr.length - 1;
  let loc = [];

  while (s < e) {
    if (arr[s] + arr[e] === x) {
      loc.push([s,e]);
      s++;
      e--;
    } else if (arr[s] + arr[e] < x) {
      s++;
    } else {
      e--;
    }
  }

  return loc;
};

console.log(twosum([1, 2, 3, 4, 5, 7, 8], 9));
console.log(twosum([2, 7, 11, 15], 9));

The step-by-step breakdown of this algorithm:

1.   Set the initial value of 's' to 0.
2.   Set the final value of 'e' to the last index (arr.length - 1).
3.   While 's' is less than 'e' meaning they have not crossed each other yet:
4.   If arr[s] + arr[e] equals x, we have found a match.
4.1. Increment 's' by 1 as there are no possible combinations before the current 's'.
4.2. Decrement 'e' by 1 as there are no possible combinations after the current 'e'.
4.3. Save the indexes where the match was located.
5.   If arr[s] + arr[e] is less than x:
5.1. Increment 's' since there are no possible combinations before the current 's', but the possibility of a match still exists for 'e'.
6.   If arr[s] + arr[e] is greater than x:
6.1. Decrement 'e' as there are no possible combinations after the current 'e', but the possibility of a match still exists for 's'.

Improving Efficiency with Nested Loop Restructuring

Answer №1

Answer №2

Answer №3

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