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

• Last Updated : 12 Jul, 2021

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]]++;``    ``}` `    ``// Initialize 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;` `    ``cout << findX(N, M, ar1, ar2) << ``'\n'``;` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to find minimum value of X``static` `int` `findX(``int` `n, ``int` `m,``                 ``int` `ar1[], ``int` `ar2[])``{``    ` `    ``// Create a frequency array for B[]``    ``int` `freq2[] = ``new` `int` `[m];``    ``for``(``int` `i = ``0``; i < m; i++)``        ``freq2[i] = ``0``;``        ` `    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``        ``freq2[ar2[i]]++;``    ``}` `    ``// Initialize x = -1``    ``int` `x = -``1``;` `    ``// Loop from [0 to m-1]``    ``for``(``int` `i = ``0``; i < m; i++)``    ``{``        ``int` `cnt = ``0``;``        ``int` `freq1[] = ``new` `int` `[m];``        ` `        ``for``(``int` `j = ``0``; j < m; j++)``        ``{``            ``freq1[j] = ``0``;``        ``}``        ` `        ``// Create a frequency array``        ``// for fixed x for all ar[i]``        ``for``(``int` `j = ``0``; j < n; j++)``        ``{``            ``freq1[(ar1[j] + i) % m]++;``        ``}` `        ``boolean` `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``public` `static` `void` `main(String[] args)``{``    ` `    ``// 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 = ar1.length;``    ` `    ``System.out.println(findX(N, M, ar1, ar2));``}``}` `// This code is contributed by Stream_Cipher`

## 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``  ` `    ``# Initialize 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`

## C#

 `// C# program for the above approach``using` `System;``class` `GFG``{` `  ``// Function to find minimum value of X``  ``static` `int` `findX(``int` `n, ``int` `m,``int` `[]ar1 ,``int` `[]ar2)``  ``{` `    ``// Create a frequency array for B[]``    ``int` `[]freq2 = ``new` `int` `[m];``    ``for``(``int` `i = 0; i < m; i++)``      ``freq2[i] = 0;` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``      ``freq2[ar2[i]]++;``    ``}` `    ``// Initialize x = -1``    ``int` `x = -1;` `    ``// Loop from [0 to m-1]``    ``for``(``int` `i = 0; i < m; i++)``    ``{``      ``int` `cnt = 0;``      ``int` `[]freq1 = ``new` `int` `[m];       ``      ``for``(``int` `j = 0; j < m; j++)``      ``{``        ``freq1[j] = 0;``      ``}` `      ``// Create a frequency array``      ``// for fixed x for all ar[i]``      ``for``(``int` `j = 0; j < n; j++)``      ``{``        ``freq1[(ar1[j] + i) % m]++;``      ``}` `      ``Boolean 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``  ``public` `static` `void` `Main(String[] args)``  ``{` `    ``// 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 = ar1.Length;``    ``Console.Write(findX(N, M, ar1, ar2));``  ``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``
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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