Given an integer N (N ≥ 5) Then assume you might have two infinite arrays X and Y the place X[] is an array of ingredient N and every ingredient of Y[] is 2^{i} the place i is the index of the array, the duty is to search out two indices let’s say A and B that are the most worth of the index at which the prefix sum in X[] is at the very least the identical because the prefix sum of Y[] and the index worth at which the X[i] – Y[i] is most respectively, The place The worth of B must be lower than or equal to A. Formally, B ≤ A.
Examples:
Enter: N = 7
Output: A = 5, B = 3
Rationalization: Y[] = {1, 2, 4, 8….., 2i – n}, X[] = {7, 7, 7, 7, …..7}
The sum of X[] until index 5 is: 35
The sum of Y[] until index 5 is: 31
5 is the utmost worth of index at which, The sum of components of X[] is strictly larger than or equal to Y[]. Most attainable distinction of sum(Xi – Yi) is at index 3, Which is 14. Due to this fact, A = 5 and B = 3.Enter: N = 6
Output: A = 4, B = 3
Rationalization: It may be verified that 4 is most attainable worth of index at which sum of components of X[] is strictly larger than or equal to Y[] and max distinction is at index 3. Due to this fact, A = 4 and B = 3.
Method: Implement the thought under to resolve the issue:
Take two lengthy knowledge kind integers shall we say Okay and L to retailer sum of X and Y respectively, run some time loop until the sum distinction is bigger than or equal to zero.
Observe the steps to resolve the issue:
 Take two lengthy knowledge kind variables let’s say Okay and L to retailer the sum of X[] and Y[] respectively until index i, different two integers A and B for output as mentioned in the issue assertion.
 Take one other variable Z, and initialize it to 0, Which can use to retailer the distinction at index i as X[i] – Y[i]. Run a Whereas loop until Z ≥ 0, and observe the under steps contained in the loop :
 Replace the worth of Okay and L.
 Replace the worth of Z as Xi – Yi.
 Increment A.
 If Z is the utmost present worth of the index at i, Then replace B as i.
 Print the worth of A and B.
Beneath is the code to implement the above strategy:
Java

Time Complexity: O(A)
Auxiliary Area: O(1)