# Minimum value by which each Array element must be added as per given conditions

Given 2 arrays A[] and B[] and an integer M. The task is to find the minimum value of X such that after changing all the elements of the array to (arr[i] + X)%M frequency of all the elements of A[] is same as the frequency of all the elements of B[]. Print “-1” if it is not possible to find any value of X.
Examples:

Input: M = 3, A[] = {0, 0, 2, 1}, B[] = {2, 0, 1, 1}
Output:
Explanation:
On taking x = 1 the numbers are changed to:
(0 + 1) % 3 = 1
(0 + 1) % 3 = 1
(2 + 1) % 3 = 0
(1 + 1) % 3 = 2
Hence on rearranging 1, 1, 0, 2 to 2, 0, 1, 1, array B[] is obtained.
Input: M = 887, A[] = {4625, 5469, 2038, 5916}, B[] = {744, 211, 795, 695}
Output: -1
Explanation:
The conversion of A[] to B[] is not possible.

Approach: The possible value of X will be in the range [0, M] as the value after the range M will give the same result we are performing modulo to M. Below are the steps:

1. Create the frequency array(say freqB[]) of the array B[].
2. Now, iterate for all possible value of X in the range [0, M] and do the following:
• For each value of X in the above range update the array A[] to (arr[i] + X)%M.
• Create the frequency array(say freqA[]) of the array A[].
• If the frequency of the array freqA[] and freqB[] is the same then print this value of X.
• Else check for another value of X.
3. After the above step if we don’t find the value of X then print “-1”.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find minimum value of X ` `int` `findX(``int` `n, ``int` `m, ` `          ``int` `ar1[], ``int` `ar2[]) ` `{ ` `    ``// Create a frequency array for B[] ` `    ``int` `freq2[m] = { 0 }; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``freq2[ar2[i]]++; ` `    ``} ` ` `  `    ``// Intialize x = -1 ` `    ``int` `x = -1; ` ` `  `    ``// Loop from [0 to m-1] ` `    ``for` `(``int` `i = 0; i < m; i++) { ` ` `  `        ``int` `cnt = 0; ` `        ``int` `freq1[m] = { 0 }; ` ` `  `        ``// Create a frequency array ` `        ``// for fixed x for all ar[i] ` `        ``for` `(``int` `j = 0; j < n; j++) { ` ` `  `            ``freq1[(ar1[j] + i) % m]++; ` `        ``} ` ` `  `        ``bool` `flag = ``true``; ` ` `  `        ``// Comparing freq1[] and freq2[] ` `        ``for` `(``int` `k = 0; k < m; k++) { ` ` `  `            ``if` `(freq1[k] != freq2[k]) { ` `                ``flag = ``false``; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// If condition is satisfied ` `        ``// then break out from loop ` `        ``if` `(flag) { ` `            ``x = i; ` `            ``break``; ` `        ``} ` `    ``} ` ` `  `    ``// Return the answer ` `    ``return` `x; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given value of M ` `    ``int` `M = 3; ` ` `  `    ``// Given arrays ar1[] and ar2[] ` `    ``int` `ar1[] = { 0, 0, 2, 1 }; ` `    ``int` `ar2[] = { 2, 0, 1, 1 }; ` ` `  `    ``int` `N = ``sizeof` `arr1 / ``sizeof` `arr1[0]; ` ` `  `    ``cout << findX(N, M, ar1, ar2) << ``'\n'``; ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 program for  ` `# the above approach ` ` `  `# Function to find  ` `# minimum value of X ` `def` `findX(n, m,  ` `          ``ar1, ar2): ` ` `  `    ``# Create a frequency  ` `    ``# array for B[] ` `    ``freq2 ``=` `[``0``] ``*` `m ` ` `  `    ``for` `i ``in` `range``(n): ` `        ``freq2[ar2[i]] ``+``=` `1` `   `  `    ``# Intialize x = -1 ` `    ``x ``=` `-``1` ` `  `    ``# Loop from [0 to m - 1] ` `    ``for` `i ``in` `range``(m): ` `        ``cnt ``=` `0` `        ``freq1 ``=` `[``0``] ``*` `m ` ` `  `        ``# Create a frequency array ` `        ``# for fixed x for all ar[i] ` `        ``for` `j ``in` `range``(n): ` `            ``freq1[(ar1[j] ``+` `i) ``%` `m] ``+``=` `1` `        `  `        ``flag ``=` `True` ` `  `        ``# Comparing freq1[]  ` `        ``# and freq2[] ` `        ``for` `k ``in` `range``(m): ` `            ``if` `(freq1[k] !``=` `freq2[k]): ` `                ``flag ``=` `False` `                ``break` `      `  `        ``# If condition is satisfied ` `        ``# then break out from loop ` `        ``if` `(flag): ` `            ``x ``=` `i ` `            ``break` ` `  `    ``# Return the answer ` `    ``return` `x ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` `   `  `    ``# Given value of M ` `    ``M ``=` `3` ` `  `    ``# Given arrays ar1[]  ` `    ``# and ar2[] ` `    ``ar1 ``=` `[``0``, ``0``, ``2``, ``1``] ` `    ``ar2 ``=` `[``2``, ``0``, ``1``, ``1``] ` ` `  `    ``N ``=` `len``(ar1) ` `    ``print` `(findX(N, M, ar1, ar2)) ` `     `  `# This code is contributed by Chitranayal`

Output:

```1
```

Time Complexity: O(N*M), where N is the number of elements in the array and M is an integer.
Auxiliary Space: O(N)

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