Minimum number of subtract operation to make an array decreasing - GeeksforGeeks

14 Apr, 2021
You are given a sequence of numbers arr[0], arr[1], ..., arr[N – 1] and a positive integer K. In each operation, you may subtract K from any element of the array. You are required to find the minimum number of operations to make the given array decreasing.An array is called decreasing if for each i: .  
Input : N = 4, K = 5, arr[] = {1, 1, 2, 3} Output : 3Explanation : Since arr[1] == arr[0] so no subtraction is required for arr[1]. For arr[2], since arr[2] > arr[1] (2 > 1) so we have to subtract arr[2] by k and after the one subtraction value of arr[2] is -3 which is less than the value of arr[1], so number of subtraction required only 1 and now value of arr[2] has been updated by -3. Similarly for arr[3], since arr[3] > arr[2] (3 > -3) so for this we have to subtract arr[3] by k two times to make the value of arr[3] lesser than arr[2], the number of subtraction required 2 and the updated value of arr[3] is -7. Now count total number of subtraction /operation required by adding number of operation on each step and that is = 0+1+2 = 3. Input : N = 5, K = 2, arr[] = {5, 4, 3, 2, 1} Output : 0
 
Approach :  
1. Traverse each element of array from 1 to n-1.
2. Check if (arr[i] > arr[i-1]) then
    Find noOfSubtraction; 
     noOfSubtraction = 
If ( (arr[i] - arr[i-1]) % k == 0 )
then noOfSubtraction++
Modify arr[i]; 
arr[i] = 
Below is implementation of above approach :  
CPP
Java
Python3
C#
PHP
Javascript
// CPP program to make an array decreasing#include <bits/stdc++.h>using namespace std; // Function to count minimum no of operationint min_noOf_operation(int arr[], int n, int k){    int noOfSubtraction;    int res = 0;    for (int i = 1; i < n; i++) {        noOfSubtraction = 0;         if (arr[i] > arr[i - 1]) {             // Count how many times we have to subtract.            noOfSubtraction = (arr[i] - arr[i - 1]) / k;             // Check an additional subtraction is            // required or not.            if ((arr[i] - arr[i - 1]) % k != 0)                noOfSubtraction++;             // Modify the value of arr[i].            arr[i] = arr[i] - k * noOfSubtraction;        }         // Count total no of operation/subtraction .        res = res + noOfSubtraction;    }     return res;} // Driver Codeint main(){    int arr[] = { 1, 1, 2, 3 };    int N = sizeof(arr) / sizeof(arr[0]);    int k = 5;    cout << min_noOf_operation(arr, N, k) << endl;    return 0;}
// Java program to make an// array decreasingimport java.util.*;import java.lang.*; public class GfG{         // Function to count minimum no of operation    public static int min_noOf_operation(int arr[],                                      int n, int k)    {        int noOfSubtraction;        int res = 0;                 for (int i = 1; i < n; i++) {            noOfSubtraction = 0;             if (arr[i] > arr[i - 1]) {                     // Count how many times                // we have to subtract.                noOfSubtraction = (arr[i] - arr[i - 1]) / k;                 // Check an additional subtraction                // is required or not.                if ((arr[i] - arr[i - 1]) % k != 0)                    noOfSubtraction++;                 // Modify the value of arr[i]                arr[i] = arr[i] - k * noOfSubtraction;            }             // Count total no of subtraction            res = res + noOfSubtraction;        }         return res;    }         // driver function    public static void main(String argc[]){        int arr = { 1, 1, 2, 3 };        int N = 4;        int k = 5;        System.out.println(min_noOf_operation(arr,                                           N, k));    }     } /* This code is contributed by Sagar Shukla */
# Python program to make an array decreasing # Function to count minimum no of operationdef min_noOf_operation(arr, n, k):     res = 0    for i in range(1,n):        noOfSubtraction = 0         if (arr[i] > arr[i - 1]):             # Count how many times we have to subtract.            noOfSubtraction = (arr[i] - arr[i - 1]) / k;             # Check an additional subtraction is            # required or not.            if ((arr[i] - arr[i - 1]) % k != 0):                noOfSubtraction+=1             # Modify the value of arr[i].            arr[i] = arr[i] - k * noOfSubtraction                  # Count total no of operation/subtraction .        res = res + noOfSubtraction          return int(res)  # Driver Codearr = [ 1, 1, 2, 3 ]N = len(arr)k = 5print(min_noOf_operation(arr, N, k)) # This code is contributed by# Smitha Dinesh Semwal
// C# program to make an// array decreasingusing System; public class GfG{         // Function to count minimum no of operation    public static int min_noOf_operation(int []arr,                                       int n, int k)    {        int noOfSubtraction;        int res = 0;                 for (int i = 1; i < n; i++) {            noOfSubtraction = 0;             if (arr[i] > arr[i - 1]) {                     // Count how many times                // we have to subtract.                noOfSubtraction = (arr[i] - arr[i - 1]) / k;                 // Check an additional subtraction                // is required or not.                if ((arr[i] - arr[i - 1]) % k != 0)                    noOfSubtraction++;                 // Modify the value of arr[i]                arr[i] = arr[i] - k * noOfSubtraction;            }             // Count total no of subtraction            res = res + noOfSubtraction;        }         return res;    }         // driver function    public static void Main()    {        int []arr = { 1, 1, 2, 3 };        int N = 4;        int k = 5;        Console.WriteLine(min_noOf_operation(arr,                                        N, k));    }     } // This code is contributed by vt_m
<?php// PHP program to make an array decreasing // Function to count minimum no of operationfunction min_noOf_operation($arr, $n, $k){         $noOfSubtraction;    $res = 0;    for($i = 1; $i < $n; $i++)    {        $noOfSubtraction = 0;         if ($arr[$i] > $arr[$i - 1])        {             // Count how many times we            // have to subtract.            $noOfSubtraction = ($arr[$i] -                      $arr[$i - 1]) / $k;             // Check an additional subtraction            // is required or not.            if (($arr[$i] - $arr[$i - 1])                               % $k != 0)                $noOfSubtraction++;             // Modify the value of arr[i].            $arr[$i] = $arr[$i] - $k *                      $noOfSubtraction;        }         // Count total no of        // operation/subtraction .        $res = $res + $noOfSubtraction;    }     return floor($res);}     // Driver Code    $arr = array(1, 1, 2, 3);    $N = count($arr);    $k = 5;    echo min_noOf_operation($arr, $N, $k) ; // This code is contributed by anuj_67.?>
<script> // JavaScript program to make an// array decreasing     // Function to count minimum no of operation    function min_noOf_operation(arr, n, k)    {        let noOfSubtraction;        let res = 0;                   for (let i = 1; i < n; i++)        {            noOfSubtraction = 0;               if (arr[i] > arr[i - 1])            {                       // Count how many times                // we have to subtract.                noOfSubtraction = (arr[i] - arr[i - 1]) / k;                   // Check an additional subtraction                // is required or not.                if ((arr[i] - arr[i - 1]) % k != 0)                    noOfSubtraction++;                   // Modify the value of arr[i]                arr[i] = arr[i] - k * noOfSubtraction;            }               // Count total no of subtraction            res = res + noOfSubtraction;        }        return res;    } // Driver code           let arr = [ 1, 1, 2, 3 ];        let N = 4;        let k = 5;        document.write(Math.floor(min_noOf_operation(arr,                                           N, k)));                                                   // This code is contributed by code_hunt.              </script>
Output :  
 3
Time Complexity : O(N). 
vt_m
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