# Find sub-arrays from given two arrays such that they have equal sum

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 = 6

Input: A[] = {10, 1}, B[] = {5, 3}
Output: -1
No such sub-arrays.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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 = 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); ` ` `  `    ``// 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 `

Output:

```Indices in array 1 : 0, 1, 2
Indices in array 2 : 0
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : rituraj_jain