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# Minimum range increment operations to Sort an array

Given an array containing N elements. It is allowed to do the below move any number of times on the array:

• Choose any L and R and increment all numbers in range L to R by 1.

The task is to find the minimum number of such moves required to sort the array in non decreasing order.

Examples:

```Input : arr[] = {1, 2, 3, 4}
Output : 0

Input : arr[] = {3, 2, 1}
Output : 2
Step 1: L=1 and R=2 (0-based)
Step 2: L=2 and R=2
Resultant array [3, 3, 3]```

Considering a sorted array, incrementing all elements of the array would still result in a sorted array.

So the idea is to traverse the elements of the array from right starting from the last index and keeping track of the minimum element. If at any point, the order of element is found to be increasing calculate the number of moves by subtracting the min element on right from current element.

Below is the implementation of the above approach:

## C++

 `// C++ program to find minimum range``// increments to sort an array` `#include ``using` `namespace` `std;` `// Function to find minimum range``// increments to sort an array``int` `minMovesToSort(``int` `arr[], ``int` `n)``{``    ``int` `moves = 0;` `    ``int` `i, mn = arr[n - 1];` `    ``for` `(i = n - 2; i >= 0; i--) {` `        ``// If current element is found greater than``        ``// last element``        ``// Increment all terms in``        ``// range i+1 to n-1``        ``if` `(arr[i] > mn)``            ``moves += arr[i] - mn;` `        ``mn = arr[i]; ``// Minimum in range i to n-1``    ``}` `    ``return` `moves;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { 3, 5, 2, 8, 4 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << minMovesToSort(arr, n);` `    ``return` `0;``}`

## Java

 `// Java program to find minimum range``// increments to sort an array`  `import` `java.io.*;` `class` `GFG {` `// Function to find minimum range``// increments to sort an array``static` `int` `minMovesToSort(``int` `arr[], ``int` `n)``{``    ``int` `moves = ``0``;` `    ``int` `i, mn = arr[n - ``1``];` `    ``for` `(i = n - ``2``; i >= ``0``; i--) {` `        ``// If current element is found greater than``        ``// last element``        ``// Increment all terms in``        ``// range i+1 to n-1``        ``if` `(arr[i] > mn)``            ``moves += arr[i] - mn;` `        ``mn = arr[i]; ``// Minimum in range i to n-1``    ``}` `    ``return` `moves;``}` `// Driver Code`  `    ``public` `static` `void` `main (String[] args) {``        ``int` `arr[] = { ``3``, ``5``, ``2``, ``8``, ``4` `};` `    ``int` `n = arr.length;` `    ``System.out.println( minMovesToSort(arr, n));``    ``}``}``// This code is contributed by anuj_67..`

## Python3

 `# Python3 program to find minimum range``# increments to sort an array` `# Function to find minimum range``# increments to sort an array``def` `minMovesToSort(arr, n) :` `    ``moves ``=` `0` `    ``mn ``=` `arr[n ``-` `1``]``    ` `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``) :` `        ``# If current element is found``        ``# greater than last element``        ``# Increment all terms in``        ``# range i+1 to n-1``        ``if` `(arr[i] > mn) :``            ``moves ``+``=` `arr[i] ``-` `mn` `        ``mn ``=` `arr[i] ``# Minimum in range i to n-1``    ` `    ``return` `moves` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:` `    ``arr ``=` `[ ``3``, ``5``, ``2``, ``8``, ``4` `]` `    ``n ``=` `len``(arr)` `    ``print``(minMovesToSort(arr, n))` `# This code is contributed by Ryuga`

## C#

 `// C# program to find minimum range``// increments to sort an array``using` `System;` `class` `GFG``{``// Function to find minimum range``// increments to sort an array``static` `int` `minMovesToSort(``int` `[]arr,``                          ``int` `n)``{``    ``int` `moves = 0;` `    ``int` `i, mn = arr[n - 1];` `    ``for` `(i = n - 2; i >= 0; i--)``    ``{` `        ``// If current element is found``        ``// greater than last element``        ``// Increment all terms in``        ``// range i+1 to n-1``        ``if` `(arr[i] > mn)``            ``moves += arr[i] - mn;` `        ``mn = arr[i]; ``// Minimum in range``                     ``// i to n-1``    ``}``    ``return` `moves;``}` `// Driver Code``static` `public` `void` `Main ()``{``    ``int` `[]arr = { 3, 5, 2, 8, 4 };``    ``int` `n = arr.Length;``    ` `    ``Console.WriteLine(minMovesToSort(arr, n));``}``}` `// This code is contributed by ajit`

## PHP

 `= 0; ``\$i``--)``    ``{` `        ``// If current element is found``        ``// greater than last element``        ``// Increment all terms in``        ``// range i+1 to n-1``        ``if` `(``\$arr``[``\$i``] > ``\$mn``)``            ``\$moves` `+= ``\$arr``[``\$i``] - ``\$mn``;` `        ``\$mn` `= ``\$arr``[``\$i``]; ``// Minimum in range i to n-1``    ``}` `    ``return` `\$moves``;``}` `// Driver Code``\$arr` `= ``array``(3, 5, 2, 8, 4 );` `\$n` `= sizeof(``\$arr``);` `echo` `minMovesToSort(``\$arr``, ``\$n``);` `// This code is contributed``// by Akanksha Rai``?>`

## Javascript

 ``

Output

`7`

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

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