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Minimum increment or decrement operations required to make the array sorted
• Difficulty Level : Expert
• Last Updated : 18 Oct, 2019

Given an array arr[] of N integers, the task is to sort the array in non-decreasing order by performing the minimum number of operations. In a single operation, an element of the array can either be incremented or decremented by 1. Print the minimum number of operations required.

Examples:

Input: arr[] = {1, 2, 1, 4, 3}
Output: 2
Add 1 to the 3rd element(1) and subtract 1 from
the 4th element(4) to get {1, 2, 2, 3, 3}

Input: arr[] = {1, 2, 2, 100}
Output: 0
Given array is already sorted.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Observation: Since we would like to minimize the number of operations needed to sort the array the following should hold:

• A number will never be decreased to value lesser than the minimum of the initial array.
• A number will never be increased to a value greater than the maximum of the initial array.
• The number of operations required to change a number from X to Y is abs(X – Y).

Approach : Based on the above observation, this problem can be solved using dynamic programming.

1. Let DP(i, j) represent the minimum operations needed to make the 1st i elements of the array sorted in non-decreasing order when the ith element is equal to j.
2. Now DP(N, j) needs to be calculated for all possible values of j where N is the size of the array. According to the observations, j ≥ smallest element of the initial array and j ≤ the largest element of the initial array.
3. The base cases in the DP(i, j) where i = 1 can be easily answered. What are the minimum operations needes to sort the 1st element in non-decreasing order such that the 1st element is equal to j?. DP(1, j) = abs( array – j).
4. Now consider DP(i, j) for i > 1. If ith element is set to j then the 1st i – 1 elements need to be sorted and the (i – 1)th element has to be ≤ j i.e. DP(i, j) = (minimum of DP(i – 1, k) where k goes from 1 to j) + abs(array[i] – j)
5. Using the above recurrence relation and the base cases, the result can be easily calculated.

Below is the implementation of the above aprpoach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;`` ` `// Function to return the minimum number``// of given operations required``// to sort the array``int` `getMinimumOps(vector<``int``> ar)``{``    ``// Number of elements in the array``    ``int` `n = ar.size();`` ` `    ``// Smallest element in the array``    ``int` `small = *min_element(ar.begin(), ar.end());`` ` `    ``// Largest element in the array``    ``int` `large = *max_element(ar.begin(), ar.end());`` ` `    ``/*``        ``dp(i, j) represents the minimum number``        ``of operations needed to make the ``        ``array[0 .. i] sorted in non-decreasing``        ``order given that ith element is j``    ``*/``    ``int` `dp[n][large + 1];`` ` `    ``// Fill the dp[]][ array for base cases``    ``for` `(``int` `j = small; j <= large; j++) {``        ``dp[j] = ``abs``(ar - j);``    ``}`` ` `    ``/*``        ``Using results for the first (i - 1) ``        ``elements, calculate the result ``        ``for the ith element``    ``*/``    ``for` `(``int` `i = 1; i < n; i++) {``        ``int` `minimum = INT_MAX;``        ``for` `(``int` `j = small; j <= large; j++) {`` ` `            ``/*``            ``If the ith element is j then we can have``            ``any value from small to j for the i-1 th``            ``element``            ``We choose the one that requires the ``            ``minimum operations``        ``*/``            ``minimum = min(minimum, dp[i - 1][j]);``            ``dp[i][j] = minimum + ``abs``(ar[i] - j);``        ``}``    ``}`` ` `    ``/*``        ``If we made the (n - 1)th element equal to j``        ``we required dp(n-1, j) operations``        ``We choose the minimum among all possible ``        ``dp(n-1, j) where j goes from small to large``    ``*/``    ``int` `ans = INT_MAX;``    ``for` `(``int` `j = small; j <= large; j++) {``        ``ans = min(ans, dp[n - 1][j]);``    ``}`` ` `    ``return` `ans;``}`` ` `// Driver code``int` `main()``{``    ``vector<``int``> ar = { 1, 2, 1, 4, 3 };`` ` `    ``cout << getMinimumOps(ar);`` ` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;`` ` `class` `GFG ``{`` ` `// Function to return the minimum number``// of given operations required``// to sort the array``static` `int` `getMinimumOps(Vector ar)``{``    ``// Number of elements in the array``    ``int` `n = ar.size();`` ` `    ``// Smallest element in the array``    ``int` `small = Collections.min(ar);`` ` `    ``// Largest element in the array``    ``int` `large = Collections.max(ar);`` ` `    ``/*``        ``dp(i, j) represents the minimum number``        ``of operations needed to make the ``        ``array[0 .. i] sorted in non-decreasing``        ``order given that ith element is j``    ``*/``    ``int` `[][]dp = ``new` `int``[n][large + ``1``];`` ` `    ``// Fill the dp[]][ array for base cases``    ``for` `(``int` `j = small; j <= large; j++)``    ``{``        ``dp[``0``][j] = Math.abs(ar.get(``0``) - j);``    ``}`` ` `    ``/*``        ``Using results for the first (i - 1) ``        ``elements, calculate the result ``        ``for the ith element``    ``*/``    ``for` `(``int` `i = ``1``; i < n; i++) ``    ``{``        ``int` `minimum = Integer.MAX_VALUE;``        ``for` `(``int` `j = small; j <= large; j++)``        ``{`` ` `            ``/*``            ``If the ith element is j then we can have``            ``any value from small to j for the i-1 th``            ``element``            ``We choose the one that requires the ``            ``minimum operations``            ``*/``            ``minimum = Math.min(minimum, dp[i - ``1``][j]);``            ``dp[i][j] = minimum + Math.abs(ar.get(i) - j);``        ``}``    ``}`` ` `    ``/*``        ``If we made the (n - 1)th element equal to j``        ``we required dp(n-1, j) operations``        ``We choose the minimum among all possible ``        ``dp(n-1, j) where j goes from small to large``    ``*/``    ``int` `ans = Integer.MAX_VALUE;``    ``for` `(``int` `j = small; j <= large; j++) ``    ``{``        ``ans = Math.min(ans, dp[n - ``1``][j]);``    ``}``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``Integer []arr = { ``1``, ``2``, ``1``, ``4``, ``3` `}; ``    ``Vector ar = ``new` `Vector<>(Arrays.asList(arr));`` ` `    ``System.out.println(getMinimumOps(ar));``}``}`` ` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach`` ` `# Function to return the minimum number``# of given operations required``# to sort the array``def` `getMinimumOps(ar):``     ` `    ``# Number of elements in the array``    ``n ``=` `len``(ar)`` ` `    ``# Smallest element in the array``    ``small ``=` `min``(ar)`` ` `    ``# Largest element in the array``    ``large ``=` `max``(ar)`` ` `    ``"""``        ``dp(i, j) represents the minimum number``        ``of operations needed to make the``        ``array[0 .. i] sorted in non-decreasing``        ``order given that ith element is j``    ``"""``    ``dp ``=` `[[ ``0` `for` `i ``in` `range``(large ``+` `1``)] ``              ``for` `i ``in` `range``(n)]`` ` `    ``# Fill the dp[]][ array for base cases``    ``for` `j ``in` `range``(small, large ``+` `1``):``        ``dp[``0``][j] ``=` `abs``(ar[``0``] ``-` `j)``    ``"""``    ``/*``        ``Using results for the first (i - 1)``        ``elements, calculate the result``        ``for the ith element``    ``*/``    ``"""``    ``for` `i ``in` `range``(``1``, n):``        ``minimum ``=` `10``*``*``9``        ``for` `j ``in` `range``(small, large ``+` `1``):``             ` `        ``# """``        ``#     /*``        ``#     If the ith element is j then we can have``        ``#     any value from small to j for the i-1 th``        ``#     element``        ``#     We choose the one that requires the``        ``#     minimum operations``        ``# """``            ``minimum ``=` `min``(minimum, dp[i ``-` `1``][j])``            ``dp[i][j] ``=` `minimum ``+` `abs``(ar[i] ``-` `j)``    ``"""``    ``/*``        ``If we made the (n - 1)th element equal to j``        ``we required dp(n-1, j) operations``        ``We choose the minimum among all possible``        ``dp(n-1, j) where j goes from small to large``    ``*/``    ``"""``    ``ans ``=` `10``*``*``9``    ``for` `j ``in` `range``(small, large ``+` `1``):``        ``ans ``=` `min``(ans, dp[n ``-` `1``][j])`` ` `    ``return` `ans`` ` `# Driver code``ar ``=` `[``1``, ``2``, ``1``, ``4``, ``3``]`` ` `print``(getMinimumOps(ar))`` ` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Linq;``using` `System.Collections.Generic;             ``     ` `class` `GFG ``{`` ` `// Function to return the minimum number``// of given operations required``// to sort the array``static` `int` `getMinimumOps(List<``int``> ar)``{``    ``// Number of elements in the array``    ``int` `n = ar.Count;`` ` `    ``// Smallest element in the array``    ``int` `small = ar.Min();`` ` `    ``// Largest element in the array``    ``int` `large = ar.Max();`` ` `    ``/*``        ``dp(i, j) represents the minimum number``        ``of operations needed to make the ``        ``array[0 .. i] sorted in non-decreasing``        ``order given that ith element is j``    ``*/``    ``int` `[,]dp = ``new` `int``[n, large + 1];`` ` `    ``// Fill the dp[], array for base cases``    ``for` `(``int` `j = small; j <= large; j++)``    ``{``        ``dp[0, j] = Math.Abs(ar - j);``    ``}`` ` `    ``/*``        ``Using results for the first (i - 1) ``        ``elements, calculate the result ``        ``for the ith element``    ``*/``    ``for` `(``int` `i = 1; i < n; i++) ``    ``{``        ``int` `minimum = ``int``.MaxValue;``        ``for` `(``int` `j = small; j <= large; j++)``        ``{`` ` `            ``/*``            ``If the ith element is j then we can have``            ``any value from small to j for the i-1 th``            ``element``            ``We choose the one that requires the ``            ``minimum operations``            ``*/``            ``minimum = Math.Min(minimum, dp[i - 1, j]);``            ``dp[i, j] = minimum + Math.Abs(ar[i] - j);``        ``}``    ``}`` ` `    ``/*``        ``If we made the (n - 1)th element equal to j``        ``we required dp(n-1, j) operations``        ``We choose the minimum among all possible ``        ``dp(n-1, j) where j goes from small to large``    ``*/``    ``int` `ans = ``int``.MaxValue;``    ``for` `(``int` `j = small; j <= large; j++) ``    ``{``        ``ans = Math.Min(ans, dp[n - 1, j]);``    ``}``    ``return` `ans;``}`` ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]arr = { 1, 2, 1, 4, 3 }; ``    ``List<``int``> ar = ``new` `List<``int``>(arr);`` ` `    ``Console.WriteLine(getMinimumOps(ar));``}``}`` ` `// This code is contributed by 29AjayKumar`
Output:
```2
```

Complexity Analysis: Time complexity for the above approach is O(N * R) where N is the number of elements in the array and R = largest – smallest element of the array + 1.

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