const arr1 = [1,3,4,5]; const arr2 = [1,3,4,4]; const asum = 9 // neive approach const isSumAvailable1 = (arr,sum) => { for (i=0; i<arr.length;i++){ for (j=i+1; j<arr.length;j++){ if(arr[j]+arr[i] == sum) return true; } } return false; } const isSumAbailable2 = (arr,sum) => { const aSet = new Set(); for (i=0; i<arr.length;i++){ if(aSet.has(arr[i])) return true; aSet.add(sum - arr[i]) } return false } console.log(isSumAbailable2(arr1,asum)); console.log(isSumAbailable2(arr2,asum));