Solution that i got , not a good one but at-least mine ๐ , the Time Complexity is O(n^2) which is not optimal .
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
let ans
function getAllIndexes(arr, val) {
var indexes = [], i;
for(i = 0; i < arr.length; i++)
if (arr[i] === val)
indexes.push(i);
return indexes;
}
nums.forEach((item,index)=>{
const foundIndex = getAllIndexes(nums , target-item).filter((id)=> id !== index)[0];
if(foundIndex>=0 && ans === undefined){
ans = [index ,foundIndex];
return;
}
})
return ans
};
Another way of solving , using 2 pointers
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
let left = 0;
let right = 1
while(left <= nums.length-1){
if(nums[left] + nums[right] === target){
return [left,right]
}else if (right === nums.length-1){
left++;
right = left+1;
}
else{
right++;
}
}
};
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