# Minimum increments by index value required to obtain at least two equal Array elements

• Difficulty Level : Hard
• Last Updated : 22 Jun, 2021

Given a strictly decreasing array arr[] consisting of N integers, the task is to find the minimum number of operations required to make at least two array elements equal, where each operation involves increasing every array element by its index value.

Examples:

Input: arr[] = {6, 5, 1}
Output:
Explanation:
{6 + 1, 5 + 2, 1 + 3} = {7, 7, 4}
Input: arr[] = {12, 8, 4}
Output:
Explanation:
Step 1 : {12 + 1, 8 + 2, 4 + 3} = {13, 10, 7}
Step 2 : {13 + 1, 10 + 2, 7 + 3} = {14, 12, 10}
Step 3 : {15, 14, 13}
Step 4 : {16, 16, 16}

Naive approach: Follow the below steps to solve the problem:

• Check if the array already has at least two equal elements or not. If found to be true, print 0.
• Otherwise, keep updating the array by increasing each array element by its index value and increase count. Check if array has two equal elements or not.
• Print count once the array is found to be containing at least two equal elements.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to update every element``// adding to it its index value``void` `update(``int` `arr[], ``int` `N)``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``arr[i] += (i + 1);``    ``}``}` `// Function to check if at least``// two elements are equal or not``bool` `check(``int` `arr[], ``int` `N)``{``    ``bool` `f = 0;``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// Count the frequency of arr[i]``        ``int` `count = 0;``        ``for` `(``int` `j = 0; j < N; j++) {` `            ``if` `(arr[i] == arr[j]) {``                ``count++;``            ``}``        ``}` `        ``if` `(count >= 2) {``            ``f = 1;``            ``break``;``        ``}``    ``}``    ``if` `(f == 1)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to calculate the number``// of increment operations required``void` `incrementCount(``int` `arr[], ``int` `N)``{``    ``// Stores the minimum number of steps``    ``int` `min = 0;` `    ``while` `(check(arr, N) != ``true``) {``        ``update(arr, N);``        ``min++;``    ``}` `    ``cout << min;``}` `// Driver Code``int` `main()``{``    ``int` `N = 3;` `    ``int` `arr[N] = { 12, 8, 4 };` `    ``incrementCount(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{` `// Function to update every element``// adding to it its index value``static` `void` `update(``int` `arr[], ``int` `N)``{``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``arr[i] += (i + ``1``);``    ``}``}` `// Function to check if at least``// two elements are equal or not``static` `boolean` `check(``int` `arr[], ``int` `N)``{``    ``int` `f = ``0``;``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// Count the frequency of arr[i]``        ``int` `count = ``0``;``        ``for``(``int` `j = ``0``; j < N; j++)``        ``{``            ``if` `(arr[i] == arr[j])``            ``{``                ``count++;``            ``}``        ``}` `        ``if` `(count >= ``2``)``        ``{``            ``f = ``1``;``            ``break``;``        ``}``    ``}``    ``if` `(f == ``1``)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to calculate the number``// of increment operations required``static` `void` `incrementCount(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores the minimum number of steps``    ``int` `min = ``0``;` `    ``while` `(check(arr, N) != ``true``)``    ``{``        ``update(arr, N);``        ``min++;``    ``}``    ``System.out.println(min);``}``    ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``3``;``    ``int` `arr[] = { ``12``, ``8``, ``4` `};``    ` `    ``incrementCount(arr, N);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to update every element``# adding to it its index value``def` `update(arr, N):``    ` `    ``for` `i ``in` `range``(N):``        ``arr[i] ``+``=` `(i ``+` `1``);` `# Function to check if at least``# two elements are equal or not``def` `check(arr, N):``    ` `    ``f ``=` `0``;``    ``for` `i ``in` `range``(N):` `        ``# Count the frequency of arr[i]``        ``count ``=` `0``;``        ` `        ``for` `j ``in` `range``(N):``            ``if` `(arr[i] ``=``=` `arr[j]):``                ``count ``+``=` `1``;` `        ``if` `(count >``=` `2``):``            ``f ``=` `1``;``            ``break``;` `    ``if` `(f ``=``=` `1``):``        ``return` `True``;``    ``else``:``        ``return` `False``;` `# Function to calculate the number``# of increment operations required``def` `incrementCount(arr, N):``    ` `    ``# Stores the minimum number of steps``    ``min` `=` `0``;` `    ``while` `(check(arr, N) !``=` `True``):``        ``update(arr, N);``        ``min` `+``=` `1``;` `    ``print``(``min``);` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``N ``=` `3``;``    ``arr ``=` `[ ``12``, ``8``, ``4` `];` `    ``incrementCount(arr, N);` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{` `// Function to update every element``// adding to it its index value``static` `void` `update(``int` `[]arr, ``int` `N)``{``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``arr[i] += (i + 1);``    ``}``}` `// Function to check if at least``// two elements are equal or not``static` `bool` `check(``int` `[]arr, ``int` `N)``{``    ``int` `f = 0;``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ` `        ``// Count the frequency of arr[i]``        ``int` `count = 0;``        ``for``(``int` `j = 0; j < N; j++)``        ``{``            ``if` `(arr[i] == arr[j])``            ``{``                ``count++;``            ``}``        ``}` `        ``if` `(count >= 2)``        ``{``            ``f = 1;``            ``break``;``        ``}``    ``}``    ``if` `(f == 1)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to calculate the number``// of increment operations required``static` `void` `incrementCount(``int` `[]arr, ``int` `N)``{``    ` `    ``// Stores the minimum number of steps``    ``int` `min = 0;` `    ``while` `(check(arr, N) != ``true``)``    ``{``        ``update(arr, N);``        ``min++;``    ``}``    ``Console.WriteLine(min);``}``    ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 3;``    ``int` `[]arr = { 12, 8, 4 };``    ` `    ``incrementCount(arr, N);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by observing that by the given operation, the difference between any two adjacent elements reduces by 1 as the array is decreasing. Therefore, the minimum number of operations required is equal to the minimum difference between any two adjacent elements.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Function to calculate the minimum``// number of steps required``void` `incrementCount(``int` `arr[], ``int` `N)``{``    ``// Stores minimum difference``    ``int` `mini = arr - arr;` `    ``for` `(``int` `i = 2; i < N; i++) {` `        ``mini``            ``= min(mini, arr[i - 1] - arr[i]);``    ``}` `    ``cout << mini;``}` `// Driver Code``int` `main()``{``    ``int` `N = 3;``    ``int` `arr[N] = { 12, 8, 4 };``    ``incrementCount(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{``    ` `// Function to calculate the minimum``// number of steps required``static` `void` `incrementCount(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores minimum difference``    ``int` `mini = arr[``0``] - arr[``1``];` `    ``for``(``int` `i = ``2``; i < N; i++)``    ``{``        ``mini = Math.min(mini,``                        ``arr[i - ``1``] - arr[i]);``    ``}``    ``System.out.println(mini);``}   `` ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `N = ``3``;``    ``int` `arr[] = { ``12``, ``8``, ``4` `};``    ` `    ``incrementCount(arr, N);``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to calculate the minimum``# number of steps required``def` `incrementCount(arr, N):` `    ``# Stores minimum difference``    ``mini ``=` `arr[``0``] ``-` `arr[``1``]` `    ``for` `i ``in` `range``(``2``, N):``        ``mini ``=` `min``(mini,``                   ``arr[i ``-` `1``] ``-` `arr[i])` `    ``print``(mini)` `# Driver Code``N ``=` `3``arr ``=` `[ ``12``, ``8``, ``4` `]` `# Function call``incrementCount(arr, N)` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement``// the above approach``using` `System;` `class` `GFG{``    ` `// Function to calculate the minimum``// number of steps required``static` `void` `incrementCount(``int` `[]arr, ``int` `N)``{``    ` `    ``// Stores minimum difference``    ``int` `mini = arr - arr;` `    ``for``(``int` `i = 2; i < N; i++)``    ``{``        ``mini = Math.Min(mini,``                        ``arr[i - 1] - arr[i]);``    ``}``    ``Console.WriteLine(mini);``}    ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 3;``    ``int` `[]arr = { 12, 8, 4 };``    ` `    ``incrementCount(arr, N);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`4`

Time Complexity: O(N)
Auxiliary Space: O(1)

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