Minimum increments to make the array non-decreasing

Given an array A[] of size N along with an integer M such that 0 <= A[i] < M, the task is to output minimum number of operations required to sort A[] to non-decreasing order using following operation, choose an element let say A[i] and update it as (A[i] + 1) mod M.

Examples:

Input: N = 4, M = 5, A[] ={4, 1, 3, 2}
Output: 2
Explanation: The operations are performed as:

• Choose element A[0] = 4 and then update it as (A[0] + 1) mod 5 = (4 + 1) mod 5 = 0. Updated A[] = {0, 1, 3, 2}
• Choose element A[4] = 2 and then update it as (A[4] + 1) mod 5 = (2 + 1) mod 5 = 0. Updated A[] = {0, 1, 3, 3}

Now, it is visible that A[] is in non-decreasing order. Therefore, minimum number of operations required are 2.

Input: N = 5, M = 10, A = {0, 0, 1, 3, 7}
Output: 0
Explanation: A[] is already in non-decreasing order. Therefore, 0 operations are required.

Approach: To solve the problem, follow the below idea:

The problem can be solved using Dynamic Programming as at every index we have 2 choices to either apply the operation or don't apply the operation. We can make a check by using previous value, if the current value is greater than the previous value, then we can explore both the cases of applying and not applying the operation. Else if the current value is less than the previous value, then we have to apply the operation. After exploring all the cases, choose the minimum of them.

Step-by-step algorithm:

• Create a 2D array, say DP[][], with all elements initialize to -1.
• Define a recursive function, Rec(curr_ele, Prev, A, M) to explore both the cases when we apply the operation and increment the current value and also when we don't apply the operation.
• Declare a variable let say minOperations to store the minimum number of operations.
• Explore the first case when we apply the operation on the current index: (Prev - A[curr_index] + M) % M + Rec(curr_index + 1, Prev, A, M)
• Explore the second case when we don't apply the operation: Rec(curr_index + 1, A[curr_index], A, M
• Store the min of both the cases in minOperations.
• Return DP[curr_index][Prev] as minOperations.

Below is the implementation of the algorithm:

// C++ code to implement the approach
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

// Intialising DP array
int DP[1000][1000];

// Recursive Function to find minimum operations
int Rec(int i, int prev, vector<int>& A, int M)
{

    // If we reach at the end of array
    if (i == A.size())
        return 0;

    // If we have cover up already this value
    if (DP[i][prev] != -1)
        return DP[i][prev];

    // Taking the difference as steps
    int minOperations
        = (prev - A[i] + M) % M + Rec(i + 1, prev, A, M);

    // Check if prev number is smaller or equal
    if (A[i] >= prev)
        minOperations
            = min(Rec(i + 1, A[i], A, M), minOperations);

    // Return and update steps
    return DP[i][prev] = minOperations;
}

// Fuction to initialize DP array and iterate on it
int minOperations(int N, int M, vector<int>& A)
{

    // Fill all the values with -1
    memset(DP, -1, sizeof(DP));

    // Call Iterator function
    return Rec(0, 0, A, M);
}

// Driver code
int main()
{

    // Inputs
    int N = 3;
    int M = 2;
    vector<int> A = { 4, 3, 2 };

    // Function call
    cout << minOperations(N, M, A);
    return 0;
}

// Java program for the above approach
import java.util.Arrays;
import java.util.Vector;

public class GFG {

    // Intialising DP array
    static int[][] DP = new int[1000][1000];

    // Recursive Function to find minimum operations
    static int Rec(int i, int prev, Vector<Integer> A,
                   int M)
    {

        // If we reach at the end of array
        if (i == A.size())
            return 0;

        // If we have cover up already this value
        if (DP[i][prev] != -1)
            return DP[i][prev];

        // Taking the difference as steps
        int minOperations = (prev - A.get(i) + M) % M
                            + Rec(i + 1, prev, A, M);

        // Check if prev number is smaller or equal
        if (A.get(i) >= prev)
            minOperations = Math.min(
                Rec(i + 1, A.get(i), A, M), minOperations);

        // Return and update steps
        return DP[i][prev] = minOperations;
    }

    // Function to initialize DP array and iterate on it
    static int minOperations(int N, int M,
                             Vector<Integer> A)
    {

        // Fill all the values with -1
        for (int[] row : DP)
            Arrays.fill(row, -1);

        // Call Iterator function
        return Rec(0, 0, A, M);
    }

    // Driver code
    public static void main(String[] args)
    {

        // Inputs
        int N = 3;
        int M = 2;
        Vector<Integer> A
            = new Vector<>(Arrays.asList(4, 3, 2));

        // Function call
        System.out.println(minOperations(N, M, A));
    }
}

// This code is contributed by Susobhan Akhuli

using System;

class Program
{
    // Initializing DP array
    static int[,] DP = new int[1000, 1000];

    // Recursive Function to find minimum operations
    static int Rec(int i, int prev, int[] A, int M)
    {
        // If we reach at the end of array
        if (i == A.Length)
            return 0;

        // If we have covered this value already
        if (DP[i, prev] != -1)
            return DP[i, prev];

        // Taking the difference as steps
        int minOperations = (prev - A[i] + M) % M + Rec(i + 1, prev, A, M);

        // Check if prev number is smaller or equal
        if (A[i] >= prev)
            minOperations = Math.Min(Rec(i + 1, A[i], A, M), minOperations);

        // Return and update steps
        return DP[i, prev] = minOperations;
    }

    // Function to initialize DP array and iterate on it
    static int MinOperations(int N, int M, int[] A)
    {
        // Fill all the values with -1
        for (int i = 0; i < 1000; i++)
        {
            for (int j = 0; j < 1000; j++)
            {
                DP[i, j] = -1;
            }
        }

        // Call Iterator function
        return Rec(0, 0, A, M);
    }

    // Driver code
    static void Main()
    {
        // Inputs
        int N = 3;
        int M = 2;
        int[] A = { 4, 3, 2 };

        // Function call
        Console.WriteLine(MinOperations(N, M, A));
    }
}

// This code is contributed by rambabuguphka

// Javascript program for the above approach

// Recursive Function to find minimum operations
function rec(i, prev, A, M, DP) {
    // If we reach at the end of array
    if (i === A.length)
        return 0;

    // If we have covered this value already
    if (DP[i][prev] !== -1)
        return DP[i][prev];

    // Taking the difference as steps
    let minOperations = (prev - A[i] + M) % M
        + rec(i + 1, prev, A, M, DP);

    // Check if previous number is smaller or equal
    if (A[i] >= prev)
        minOperations = Math.min(
            rec(i + 1, A[i], A, M, DP), minOperations);

    // Return and update steps
    DP[i][prev] = minOperations;
    return DP[i][prev];
}

// Function to initialize DP array and iterate on it
function minOperations(N, M, A) {
    // Intialising DP array
    let DP = Array.from({ length: 1000 }, () => Array(1000).fill(-1));

    // Call Iterator function
    return rec(0, 0, A, M, DP);
}

// Driver code
(function main() {
    // Inputs
    let N = 3;
    let M = 2;
    let A = [4, 3, 2];

    // Function call
    console.log(minOperations(N, M, A));
})();

// This code is contributed by Susobhan Akhuli

# Recursive function to find minimum operations
def rec(i, prev, A, M, DP):
    # If we reach at the end of array
    if i == len(A):
        return 0

    # If we have covered this value already
    if DP[i][prev] != -1:
        return DP[i][prev]

    # Taking the difference as steps
    min_operations = (prev - A[i] + M) % M + rec(i + 1, prev, A, M, DP)

    # Check if previous number is smaller or equal
    if A[i] >= prev:
        min_operations = min(rec(i + 1, A[i], A, M, DP), min_operations)

    # Return and update steps
    DP[i][prev] = min_operations
    return DP[i][prev]

# Function to initialize DP array and iterate on it
def minOperations(N, M, A):
    # Intialising DP array with -1
    DP = [[-1] * 1000 for _ in range(1000)]

    # Call Iterator function
    return rec(0, 0, A, M, DP)

# Driver code
if __name__ == "__main__":
    # Inputs
    N = 3
    M = 2
    A = [4, 3, 2]

    # Function call
    print(minOperations(N, M, A))

-1

Time Complexity: O(N*N), where N is the size of input array A[].
Auxiliary Space: O(N*N)