A simpler way to write the sum_pairs function

user_10252315 0
Viewed 85

I want someone to show me a simpler way to write my sum_pairs(arr,sum) function which return the first 2 values in arr that adds up to form sum . My code works but i think it is complex , i need someone to simplify it . So, this is my code .

function sum_pairs(ints,s){
 let arr=[];
 let arrOfIndex=[];
 for(let i=0;i<ints.length;i++){
  for(let a=0;a<ints.length;a++){
   if(a!=i){
    if(ints[i]+ints[a]==s){
     let newArr=[ints[i],ints[a]];
     let sumIndex=i+a;
     arr.push(newArr);
     arrOfIndex.push(sumIndex);
    }
   }
  } 
 }
 let sortedArray=arrOfIndex.sort((a,b)=>a-b);
 return arr[arrOfIndex.indexOf(sortedArray[0])];
}

console.log(sum_pairs([7,2,5,8,4,3],7))//[2,5]

javascript
3 Answers

You could take a single loop with a hash table for the second pair element.

It works by looking up the actual value and if found, then this value is part of a pair. In this case return the delta of the sum and the value and the value.

If not found, add a new entry to the hash table with the missing value for getting a sum.

Proceed until found or end.

function sum_pairs(ints, s) {
    var hash = Object.create(null),
        i,
        value;

    for (i = 0; i < ints.length; i++) {
        value = ints[i];
        if (hash[value]) return [s - value, value];
        hash[s - value] = true;
    }
}

console.log(sum_pairs([7, 2, 5, 8, 4, 3], 7));

All pairs (with an array without duplicates)

function allPairs(ints, s) {
    var hash = Object.create(null),
        i,
        value,
        pairs = [];

    for (i = 0; i < ints.length; i++) {
        value = ints[i];
        if (hash[value]) pairs.push([s - value, value]);
        hash[s - value] = true;
    }
    return pairs;
}

console.log(allPairs([7, 2, 5, 8, 4, 3], 7));

Finally find duplicate pairs as well :-)

function allPairs(ints, s) {
    var hash = Object.create(null),
        i,
        value,
        pairs = [];

    for (i = 0; i < ints.length; i++) {
        value = ints[i];
        if (hash[value]) {
            pairs.push([s - value, value]);
            hash[value]--;
            continue;
        }
        if (!hash[s - value]) {
            hash[s - value] = 0;
        }
        ++hash[s - value];
    }
    return pairs;
}

console.log(allPairs([4, 3, 3, 4, 7, 2, 5, 8, 3], 7));

user_14476750

You can use the following code 

function sum_pairs(ints, s) {
  let results = [];
  for (let i=0; i<ints.length; i++) {
    for (let j=i+1; j<ints.length; j++) {
      if (ints[j] === s - ints[i]) {
        results.push([ints[i], ints[j]])
      }
    }
  }
  return results;
}

This will provide you all the pairs

To return a single pair you can use following 

function sum_pairs(ints, s) {
      let results = [];
      for (let i=0; i<ints.length; i++) {
        for (let j=i+1; j<ints.length; j++) {
          if (ints[j] === s - ints[i]) {
            results.push([ints[i], ints[j]])
             return results;
          }
        }
      }
    }
user_100980740

Steps

  1. Iterate through the array
  2. Find the composition of n (n is the number in each interation) by using this equation n_composition = sum - n
  3. Search for n_composition in the array
  4. If found return [n, n_comp]. If not then continue the loop. If not found at all then return null.

let n = 0;
let n_comp = 0;
let sum_pairs = (arr, sum) => {
  for(let i = 0, len = arr.length; i < len; ++i){
    n = arr[i];
    n_comp = sum - n;
    if (arr.includes(n_comp)){
      return [n, n_comp];
    }
  }
  return null;
}

console.log(sum_pairs([7,2,5,8,4,3],7))//[2,5]

user_37013930
Related