Given two arrays A[] and B[] of equal sizes i.e. N containing integers from 1 to N. The task is to find sub-arrays from the given arrays such that they have equal sum. Print the indices of such sub-arrays. If no such sub-arrays are possible then print -1.
Examples:
Input: A[] = {1, 2, 3, 4, 5}, B[] = {6, 2, 1, 5, 4}
Output:
Indices in array 1 : 0, 1, 2
Indices in array 2 : 0
A[0..2] = 1 + 2 + 3 = 6
B[0] = 6Input: A[] = {10, 1}, B[] = {5, 3}
Output: -1
No such sub-arrays.
Approach: Let Ai denote the sum of first i elements in A and Bj denote the sum of first j elements in B. Without loss of generality we assume that An <= Bn.
Now Bn >= An >= Ai. So for each Ai we can find the smallest j such that Ai <= Bj. For each i we find the difference
Bj – Ai .
If difference is 0 then we are done as the elements from 1 to i in A and 1 to j in B have the same sum. Suppose difference is not 0.Then the difference must lie in the range [1, n-1].
Proof:
Let Bj – Ai >= n
Bj >= Ai + n
Bj-1 >= Ai (As the jth element in B can be at most n so Bj <= Bj-1 + n)
Now this is a contradiction as we had assumed that j is the smallest index
such that Bj >= Ai is j. So our assumption is wrong.
So Bj – Ai < n
Now there are n such differences(corresponding to each index) but only (n-1) possible values, so at least two indices will produce the same difference(By Pigeonhole principle). Let Aj – By = Ai – Bx. On rearranging we get Aj – Ai = By – Bx. So the required subarrays are [ i+1, j ] in A and [ x+1, y ] in B.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to print the valid indices in the array void printAns(int x, int y, int num) { cout << "Indices in array " << num << " : "; for (int i = x; i < y; ++i) { cout << i << ", "; } cout << y << "\n"; } // Function to find sub-arrays from two // different arrays with equal sum void findSubarray(int N, int a[], int b[], bool swap) { // Map to store the indices in A and B // which produce the given difference std::map<int, pair<int, int> > index; int difference; index[0] = make_pair(-1, -1); int j = 0; for (int i = 0; i < N; ++i) { // Find the smallest j such that b[j] >= a[i] while (b[j] < a[i]) { j++; } difference = b[j] - a[i]; // Difference encountered for the second time if (index.find(difference) != index.end()) { // b[j] - a[i] = b[idx.second] - a[idx.first] // b[j] - b[idx.second] = a[i] = a[idx.first] // So sub-arrays are a[idx.first+1...i] and b[idx.second+1...j] if (swap) { pair<int, int> idx = index[b[j] - a[i]]; printAns(idx.second + 1, j, 1); printAns(idx.first + 1, i, 2); } else { pair<int, int> idx = index[b[j] - a[i]]; printAns(idx.first + 1, i, 1); printAns(idx.second + 1, j, 2); } return; } // Store the indices for difference in the map index[difference] = make_pair(i, j); } cout << "-1"; } // Utility function to calculate the // cumulative sum of the array void cumulativeSum(int arr[], int n) { for (int i = 1; i < n; ++i) arr[i] += arr[i - 1]; } // Driver code int main() { int a[] = { 1, 2, 3, 4, 5 }; int b[] = { 6, 2, 1, 5, 4 }; int N = sizeof(a) / sizeof(a[0]); // Function to update the arrays // with their cumulative sum cumulativeSum(a, N); cumulativeSum(b, N); if (b[N - 1] > a[N - 1]) { findSubarray(N, a, b, false); } else { // Swap is true as a and b are swapped during // function call findSubarray(N, b, a, true); } return 0; } |
Python3
# Python3 implementation of the approach # Function to print the valid indices in the array def printAns(x, y, num): print("Indices in array", num, ":", end = " ") for i in range(x, y): print(i, end = ", ") print(y) # Function to find sub-arrays from two # different arrays with equal sum def findSubarray(N, a, b, swap): # Map to store the indices in A and B # which produce the given difference index = {} difference, j = 0, 0 index[0] = (-1, -1) for i in range(0, N): # Find the smallest j such that b[j] >= a[i] while b[j] < a[i]: j += 1 difference = b[j] - a[i] # Difference encountered for the second time if difference in index: # b[j] - a[i] = b[idx.second] - a[idx.first] # b[j] - b[idx.second] = a[i] = a[idx.first] # So sub-arrays are a[idx.first+1...i] and b[idx.second+1...j] if swap: idx = index[b[j] - a[i]] printAns(idx[1] + 1, j, 1) printAns(idx[0] + 1, i, 2) else: idx = index[b[j] - a[i]] printAns(idx[0] + 1, i, 1) printAns(idx[1] + 1, j, 2) return # Store the indices for difference in the map index[difference] = (i, j) print("-1") # Utility function to calculate the # cumulative sum of the array def cumulativeSum(arr, n): for i in range(1, n): arr[i] += arr[i - 1] # Driver code if __name__ == "__main__": a = [1, 2, 3, 4, 5] b = [6, 2, 1, 5, 4] N = len(a) # Function to update the arrays # with their cumulative sum cumulativeSum(a, N) cumulativeSum(b, N) if b[N - 1] > a[N - 1]: findSubarray(N, a, b, False) else: # Swap is true as a and b are # swapped during function call findSubarray(N, b, a, True) # This code is contributed by Rituraj Jain |
Indices in array 1 : 0, 1, 2 Indices in array 2 : 0
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Improved By : rituraj_jain
