If we carefully observe, we can notice that to get the sum of the current subarray we just need to add the current element(i.e. arr[j]) to the sum of the previous subarray i.e. arr[i….j-1]. Assume previous subarray = arr[i……j-1]
current subarray = arr[i…..j]
Sum of arr[i….j] = (sum of arr[i….j-1]) + arr[j] This is how we can remove the third loop and while moving j pointer, we can calculate the sum.

First, we will run a loop(say i) that will select every possible starting index of the subarray. The possible starting indices can vary from index 0 to index n-1(n = array size).
Inside the loop, we will run another loop(say j) that will signify the ending index as well as the current element of the subarray. For every subarray starting from the index i, the possible ending index can vary from index i to n-1(n = size of the array).
Inside loop j, we will add the current element to the sum of the previous subarray i.e. sum = sum + arr[j].
After calculating the sum, we will check if the sum is equal to the given k. If it is, we will increase the value of the count.