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
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 program to make an array decreasing #include <bits/stdc++.h> using namespace std;
// Function to count minimum no of operation 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 operation/subtraction .
res = res + noOfSubtraction;
}
return res;
} // Driver Code int 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 decreasing import 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 operation def 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 Code arr = [ 1 , 1 , 2 , 3 ]
N = len (arr)
k = 5
print (min_noOf_operation(arr, N, k))
# This code is contributed by # Smitha Dinesh Semwal |
// C# program to make an // array decreasing using 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 operation function 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).
Auxiliary Space: O(1).