# Minimum number of steps to make all elements of array equal

• Difficulty Level : Medium
• Last Updated : 05 Nov, 2021

Given two arrays A[] and B[] of the same length, the task is to find the minimum number of steps required to make all the elements of the array equal by replacing the element with the difference of the corresponding element from array B.
Note: If this is not possible print -1.
Examples:

Input: A[] = {5, 7, 10, 5, 15} B[] = {2, 2, 1, 3, 5}
Output:
Explanation:
Steps to convert Array elements equal –
Index 0: 5 is not changed, Steps Taken = 0
Index 1: 7 – (2 * 1) = 5, Steps Taken = 1
Index 2: 10 – (1 * 5) = 5, Steps Taken = 5
Index 3: 5 is not changed, Steps Taken = 0
Index 4: 15 – (5 * 2) = 5, Steps Taken = 2
Therefore, total steps required = 0 + 1 + 5 + 0 + 2 = 8

Input: A[] = {5, 6}, B[] = {4, 3}
Output: -1
Explanation:
It’s not possible to make all elements of array equal.

Approach: The idea is to find the value at which the array elements can be converged, this value will definitely lie in between 0 to the minimum value of the array. Below is the illustration of the approach:

• Initially, mark the flag as False, to check that the array is convertible or not.
• Run a while loop until the minimum of the array is greater than -1.
• Iterate over the indices of the array, that is from 0 to length – 1
• For each index, check that if the element at that index in array A is greater than the minimum of the array, If yes then update the array element A[i] as A[i] – B[i].
• Increment the number of steps by 1.
• Check that all the array elements are equal if yes then mark the flag as True and break the while loop
• Otherwise, repeat the steps above to compute the total number of steps required.
• Finally, check if the flag is True, If yes then the array is convertible.

Explanation with Example:
Given arrays be – A[] = {5, 7, 10, 5, 15}, B[] = {2, 2, 1, 3, 5}

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the``// minimum number of steps to convert``// array by subtracting the corresponding``// element from array B``#include ``using` `namespace` `std;`` ` `// Function to find the minimum steps``void` `minimumSteps(``int` `ar1[], ``int` `ar2[], ``int` `n)``{``    ` `    ``// Counter to store the steps``    ``int` `ans = 0;``     ` `    ``// Flag to check that``    ``// array is converted``    ``bool` `flag = ``true``;``     ` `    ``// Loop until the minimum of the``    ``// array is greater than -1``    ``while` `(*min_element(ar1, ar1 + n) > -1)``    ``{``        ``int` `a = *min_element(ar1, ar1 + n);``        ` `        ``// Loop to convert the array``        ``// elements by subtraction``        ``for``(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(ar1[i] != a)``            ``{``                ``ar1[i] -= ar2[i];``                ``ans += 1;``            ``}``        ``}``        ` `        ``set<``int``> s1(ar1, ar1 + n);``        ` `        ``// If the array is converted``        ``if` `(s1.size() == 1)``        ``{``            ``flag = ``false``;``            ``cout << ans << endl;``            ``break``;``        ``}``    ``}``    ` `    ``if` `(flag)``    ``{``        ``cout << -1 << endl;``    ``}``}``        ` `// Driver Code``int` `main()``{``    ``int` `n = 5;``    ``int` `ar1[] = { 5, 7, 10, 5, 15 };``    ``int` `ar2[] = { 1, 2, 1, 5, 5 };``     ` `    ``// Function call``    ``minimumSteps(ar1, ar2, n);`` ` `    ``return` `0;``}` `// This code is contributed by rutvik_56`

## Java

 `// Java implementation to find the``// minimum number of steps to convert``// array by subtracting the corresponding``// element from array B``import` `java.util.*;``class` `GFG{`` ` `// Function to find the``// minimum steps``static` `void` `minimumSteps(``int` `ar1[],``                         ``int` `ar2[],``                         ``int` `n)``{``  ``// Counter to store the steps``  ``int` `ans = ``0``;` `  ``// Flag to check that``  ``// array is converted``  ``boolean` `flag = ``true``;` `  ``// Loop until the minimum of the``  ``// array is greater than -1``  ``while` `(Arrays.stream(ar1).min().getAsInt() > -``1``)``  ``{``    ``int` `a = Arrays.stream(ar1).min().getAsInt();` `    ``// Loop to convert the array``    ``// elements by subtraction``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``      ``if` `(ar1[i] != a)``      ``{``        ``ar1[i] -= ar2[i];``        ``ans += ``1``;``      ``}``    ``}` `    ``HashSet s1 = ``new` `HashSet<>();``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``      ``s1.add(ar1[i]);``    ``}` `    ``// If the array is converted``    ``if` `(s1.size() == ``1``)``    ``{``      ``flag = ``false``;``      ``System.out.print(ans + ``"\n"``);``      ``break``;``    ``}``  ``}` `  ``if` `(flag)``  ``{``    ``System.out.print(-``1` `+ ``"\n"``);``  ``}``}``        ` `// Driver Code``public` `static` `void` `main(String[] args)``{``  ``int` `n = ``5``;``  ``int` `ar1[] = {``5``, ``7``, ``10``, ``5``, ``15``};``  ``int` `ar2[] = {``1``, ``2``, ``1``, ``5``, ``5``};` `  ``// Function call``  ``minimumSteps(ar1, ar2, n);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 implementation to find the``# minimum number of steps to convert``# array by subtracting the corresponding``# element from array B` `# Function to find the minimum steps``def` `minimumSteps(ar1, ar2, n):``    ` `    ``# Counter to store the steps``    ``ans ``=` `0``    ` `    ``# Flag to check that``    ``# array is converted``    ``flag ``=` `True``    ` `    ``# Loop until the minimum of the``    ``# array is greater than -1``    ``while` `min``(ar1)>``-``1``:``        ``a ``=` `min``(ar1)``        ` `        ``# Loop to convert the array``        ``# elements by subtraction``        ``for` `i ``in` `range``(n):``            ``if` `ar1[i]!``=` `a:``                ``ar1[i]``-``=` `ar2[i]``                ``ans``+``=` `1``                ` `        ``# If the array is converted``        ``if` `len``(``set``(ar1))``=``=` `1``:``            ``flag ``=` `False``            ``print``(ans)``            ``break``    ``if` `flag:``        ``print``(``-``1``)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``n ``=` `5``    ``ar1 ``=` `[``5``, ``7``, ``10``, ``5``, ``15``]``    ``ar2 ``=` `[``1``, ``2``, ``1``, ``5``, ``5``]``    ` `    ``# Function Call``    ``minimumSteps(ar1, ar2, n)`

## C#

 `// C# implementation to``// find the minimum number``// of steps to convert array``// by subtracting the``// corresponding element from``// array B``using` `System;``using` `System.Linq;``using` `System.Collections.Generic;``class` `GFG{`` ` `// Function to find the``// minimum steps``static` `void` `minimumSteps(``int` `[]ar1,``                         ``int` `[]ar2,``                         ``int` `n)``{``  ``// Counter to store the steps``  ``int` `ans = 0;` `  ``// Flag to check that``  ``// array is converted``  ``bool` `flag = ``true``;` `  ``// Loop until the minimum of the``  ``// array is greater than -1``  ``while` `(ar1.Min() > -1)``  ``{``    ``int` `a = ar1.Min();` `    ``// Loop to convert the array``    ``// elements by subtraction``    ``for``(``int` `i = 0; i < n; i++)``    ``{``      ``if` `(ar1[i] != a)``      ``{``        ``ar1[i] -= ar2[i];``        ``ans += 1;``      ``}``    ``}` `    ``HashSet<``int``> s1 = ``new` `HashSet<``int``>();``    ` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``      ``s1.Add(ar1[i]);``    ``}` `    ``// If the array is converted``    ``if` `(s1.Count == 1)``    ``{``      ``flag = ``false``;``      ``Console.Write(ans + ``"\n"``);``      ``break``;``    ``}``  ``}` `  ``if` `(flag)``  ``{``    ``Console.Write(-1 + ``"\n"``);``  ``}``}``        ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``  ``int` `n = 5;``  ``int` `[]ar1 = {5, 7, 10, 5, 15};``  ``int` `[]ar2 = {1, 2, 1, 5, 5};` `  ``// Function call``  ``minimumSteps(ar1, ar2, n);``}``}` `// This code is contributed by 29AjayKumar`

Output:

`8`

Performance Analysis:

• Time Complexity: O(N2 logN) As in the above approach, there are two loops to convert the array. The outer loop takes O(N) and the inner loop takes O(NlogN) as we are using set in the inner loop.
• Auxiliary Space: O(N). We are using a set that takes O(N) extra space.

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