Minimum increment or decrement operations required to make the array sorted

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. 

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 1^st i elements of the array sorted in non-decreasing order when the i^th 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 1^st element in non-decreasing order such that the 1^st element is equal to j?. DP(1, j) = abs( array[1] – j).
4. Now consider DP(i, j) for i > 1. If i^th element is set to j then the 1^st 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 approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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[0][j] = abs(ar[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 = 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<Integer> 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<Integer> 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[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 = 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

<script>

// Javascript implementation of the approach

// Function to return the minimum number
// of given operations required
// to sort the array
function getMinimumOps(ar)
{
    // Number of elements in the array
    var n = ar.length;

    // Smallest element in the array
    var small = Math.min.apply(Math,ar);

    // Largest element in the array
    var large = Math.max.apply(Math,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
    */
    var dp = new Array(n);
    var i,j;
    for(i=0;i<dp.length;i++)
        dp[i] = new Array(large+1);

    // Fill the dp[]][ array for base cases
    for (j = small; j <= large; j++) {
        dp[0][j] = Math.abs(ar[0] - j);
    }

    /*
        Using results for the first (i - 1)
        elements, calculate the result
        for the ith element
    */
    for (i = 1; i < n; i++) {
        var minimum = 2147483647;
        for (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
    */
    var ans = 2147483647;
    for (j = small; j <= large; j++) {
        ans = Math.min(ans, dp[n - 1][j]);
    }

    return ans;
}

// Driver code
    var ar = [1, 2, 1, 4, 3];

    document.write(getMinimumOps(ar));

</script>

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)