# Minimum increment or decrement operations required to make the array sorted

• Difficulty Level : Expert
• Last Updated : 30 May, 2022

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:
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:

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 needs 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`

## Javascript

 ``

Output:

`2`

Complexity Analysis:
Time Complexity: O(N*R), 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.
Auxiliary Space: O(N * large)

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