Count distinct pairs with difference equal to k and further optimize.

Question

Accepted Answer

Count distinct pairs with difference equal to k and optimize.

Use a hash set to store the elements of the array.
Iterate through the array and check if the current element + k or current element - k exists in the hash set.
Increment the count if a pair is found and add the current element to the hash set.
Return the count of distinct pairs.

Accepted Answer

3 pairs (1,3) (3,5) (2,4)

Accepted Answer

This is for sum, you can do same for subtraction

function getMaxSumOfSubarray(arr, num) {

let sum = 0;

let hash = {};

let subArray = [];

let partialSum = 0;

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

let j = i + num;

subArray = arr.slice(i, j);

partialSum = getSum(subArray);

if (partialSum > sum) {

sum = partialSum;

}

hash[partialSum] = subArray;

}

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