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 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[1] – 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


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.

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.