Two Sum non-unique elements logic

This code works for the test cases having unique elements in the array but not if the same element is present like [3,2,3] so I want something like to add 3+3 === target(when the target is 6) and to get both of their indexes. Please explain me the logic and upto this rate my approach which I used here.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

function twoSum(nums, target) {
  return nums.reduce((acc, ele, index) => {
    if (ele + nums.at(index + 1) === target) {
      acc.push(index, index + 1)
    }
    return acc
  }, [])
}

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

console.log(twoSum(nums, target))

const nums1 = [3,2,3]
const target1 = 6

console.log(twoSum(nums1, target1))

<!-- end snippet -->

We can use a map and iterate over the array, storing each element's value and index. We then check for each element if its complement is already in the map. If found, we iterate over the array of indices for the complement, and for each index, add a pair of indices to the result to handle reuse of a digit.

Time complexity still: O(n)

Due to the possible multiple results, single results are also nested. We can flatten them if you do not like it.

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

const twoSum = (nums, target) => nums.reduce((acc, ele, index) => {
  let complement = target - ele;
  if (acc.map.has(complement)) {
    for (let compIndex of acc.map.get(complement)) {
      acc.result.push([compIndex, index]);
    }
  }
  if (!acc.map.has(ele)) acc.map.set(ele, []);
  acc.map.get(ele).push(index);
  return acc;
}, { map: new Map(), result: [] }).result;


const nums = [2, 7, 11, 15]
const target = 9
console.log(twoSum(nums, target)) // Output: [0, 1]

const nums1 = [3, 2, 3]
const target1 = 6
console.log(twoSum(nums1, target1)) // Output: [0, 2]

console.log(twoSum([2, 7, 11, 7, 15, 2, 7], 9)) // [[0,1], [0,3], [0,6], [1,5], [3,5], [5,6]]

<!-- end snippet -->

This one looks more elegant but is O(n^2)

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

const twoSum = (nums, target) => nums.reduce((acc, ele, i) => {
  nums.slice(i + 1).forEach((ele2, j) => {
    if (ele + ele2 === target) acc.push([i, i + j + 1]);
  });
  return acc;
}, []);

const nums = [2, 7, 11, 15]
const target = 9
console.log(twoSum(nums, target)) // Output: [0, 1]

const nums1 = [3, 2, 3]
const target1 = 6
console.log(twoSum(nums1, target1)) // Output: [0, 2]

console.log(twoSum([2, 7, 11, 7, 15, 2, 7], 9)) // [[0,1], [0,3], [0,6], [1,5], [3,5], [5,6]]

<!-- end snippet -->

Iterate all possible combinations of 2 elements and collect their indices:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

function twoSum(nums, target) {
const result = [];
for(let i = 0; i < nums.length - 1; i++){
for(let j = i + 1; j < nums.length; j++){
if(nums[i] + nums[j] === target){
result.push([i,j]);
}
}
}
return result;
}
const log = result => console.log(JSON.stringify(result));
log(twoSum([2, 7, 11, 15], 9))
log(twoSum([2, 7, 11, 7, 15, 2, 7], 9))
log(twoSum([2, 7, 11, 15], 18))
log(twoSum([3, 2, 3], 6))

<!-- end snippet -->

An Array::reduce() version:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-js -->

function twoSum(nums, target) {
return nums.reduce((result, num, i) => {
return nums.slice(i + 1).reduce((result, num2, j) => {
num + num2 === target && result.push([i, i + j + 1]);
return result;
}, result);
}, []);
}
const log = result => console.log(JSON.stringify(result));
log(twoSum([2, 7, 11, 15], 9))
log(twoSum([2, 7, 11, 7, 15, 2, 7], 9))
log(twoSum([2, 7, 11, 15], 18))
log(twoSum([3, 2, 3], 6))

<!-- end snippet -->

Benchmark with a small set:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-html -->

<script benchmark data-count="1000000">
// @benchmark mplungjan
const twoSum2 = (nums, target) => nums.reduce((acc, ele, index) => {
let complement = target - ele;
if (acc.map.has(complement)) {
for (let compIndex of acc.map.get(complement)) {
acc.result.push([compIndex, index]);
}
}
if (!acc.map.has(ele)) acc.map.set(ele, []);
acc.map.get(ele).push(index);
return acc;
}, { map: new Map(), result: [] }).result;
// @run
[twoSum2([2, 7, 11, 15], 9),
twoSum2([2, 7, 11, 7, 15, 2, 7], 9),
twoSum2([2, 7, 11, 15], 18),
twoSum2([3, 2, 3], 6)]
// @benchmark Alexander
const twoSum = (nums, target) => {
const result = [];
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] === target) {
result.push([i, j]);
}
}
}
return result;
};
// @run
[twoSum([2, 7, 11, 15], 9),
twoSum([2, 7, 11, 7, 15, 2, 7], 9),
twoSum([2, 7, 11, 15], 18),
twoSum([3, 2, 3], 6)]
</script>
<script src="https://cdn.jsdelivr.net/gh/silentmantra/benchmark/loader.js"></script>

<!-- end snippet -->

Benchmark with a big set:

<!-- begin snippet: js hide: false console: true babel: false -->

<!-- language: lang-html -->

<script benchmark data-count="1">
const random = Array.from({length:20000}, () => Math.round(Math.random()*20));
// @benchmark mplungjan
const twoSum2 = (nums, target) => nums.reduce((acc, ele, index) => {
let complement = target - ele;
if (acc.map.has(complement)) {
for (let compIndex of acc.map.get(complement)) {
acc.result.push([compIndex, index]);
}
}
if (!acc.map.has(ele)) acc.map.set(ele, []);
acc.map.get(ele).push(index);
return acc;
}, { map: new Map(), result: [] }).result;
// @run
twoSum2(random, 9)
// @benchmark Alexander
const twoSum = (nums, target) => {
const result = [];
for (let i = 0; i < nums.length; i++) {
for (let j = i + 1; j < nums.length; j++) {
if (nums[i] + nums[j] === target) {
result.push([i, j]);
}
}
}
return result;
};
// @run
twoSum(random, 9);
</script>
<script src="https://cdn.jsdelivr.net/gh/silentmantra/benchmark/loader.js"></script>

<!-- end snippet -->