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### Two Sum (Main Thread)

Here is the interview question prompt, presented for reference.

This is a classic and very common interview problem. Given an array of integers, return the indices of the two numbers in it that add up to a specific goal number.

So let's say our 'goal' number was `10`. Our numbers to sum to it would be `3` and `7`, and we would return an array of indices `1` and `3` respectively.

``````let arr = [1, 3, 6, 7, 9];
let goal = 10;
twoSum(arr, goal);
// [1, 3]
``````

You may assume that each input would have exactly one solution. Additionally, you may not use the same element twice towards the sum. This means if given `[1, 3]` and a goal of `2`, you cannot use `1` twice and return its index twice as `[0, 0]`.

Here's the function signature to fill in:

``````function twoSum(arr, goal) {
return arr;
}
``````

AlgoDaily partner CodeTips.co.uk has kindly provided a guide on solving Two Sum in Go. Check it out for even deeper coverage of this problem.

### Constraints

• Length of the array <= `100000`
• The values of the array will be between `-1000000000` and `1000000000`
• The array can be empty
• The target sum will be between `-1000000000` and `1000000000`
• Expected time complexity : `O(n)`
• Expected space complexity : `O(n)`

You can see the full challenge with visuals at this link.

Challenges • Asked almost 2 years ago by Noobert Jake from AlgoDaily Commented on Jul 29, 2019:

This is the main discussion thread generated for Two Sum. Noobert Commented on Jul 29, 2019:

function twoSum(arr, goal) {

let results = [];

for (let i = 0; i < arr.length; i++){

if (arr.indexOf(goal - arr[i]) > 0){

results.push(i, arr.indexOf(goal - arr[i]));

break;

}

}

return results;

} Vishnu Lucky Commented on Dec 14, 2019:

private static int[] getIndices(int[] a, int k) {

HashMap map = new HashMap<>();
for(int i=0;i<a.length;i++) {
map.put(a[i], i);
}

for(int i=0;i<a.length;i++) {
if(map.containsKey(k-a[i])) {
return new int[] {i,map.get(k-a[i])};
}
}

return new int[] {-1,-1};

} kcoddington Commented on Sep 17, 2020:

The following solution shows being faster than the provided solution on jsbench.me:

``````function twoSum(arr, goal) {
const results = [];
const hashmap = {};
for (let i = 0; i < arr.length; i++) {
hashmap[arr[i]] = i;
const mate = goal - arr[i];
if (hashmap[mate] != null) {
results.push(hashmap[mate]);
results.push(i);
break;
}
}
return results;
}
`````` TimWong17 Commented on Jan 17, 2021:

Is there a difference in using the Two Pointer approach versus a Hash Map for this problem? I was initially using the Two Pointer approach to solve this. I believe there time complexities are both O(n). Just not too sure which approach would be best in an interview setting. Jake from AlgoDaily Commented on Jan 19, 2021:

Hi Tim,

Both approaches work well! The hash map takes `O(n)` more space (since it needs to store each input element in the array as keys), so complexity-wise Two Pointers are better (`O(1)` space). anil77k Commented on Jan 25, 2021:

The below solution seems to work fine. But when I submit the tests are being called for sortedtwosum.

``````public static int[] twoSum(int[] arr, int goal){
int[] result = new int;
HashMap<Integer,Integer> resultMap = new HashMap<Integer,Integer>();
for(int i=0;i<arr.length;i++){
if(resultMap.containsKey(arr[i])){
result = resultMap.get(arr[i]);
result = i;
return result;
}else{
resultMap.put(goal - arr[i], i);
}
}
return result;
}
`````` Jake from AlgoDaily Commented on Jan 25, 2021:

Thank you, Anil! This is resolved.