Minimum range increment operations to Sort an array
Given an array containing N elements. It is allowed to do the below move any number of times on the array:
- Choose any L and R and increment all numbers in range L to R by 1.
The task is to find the minimum number of such moves required to sort the array in non decreasing order.
Examples:
Input : arr[] = {1, 2, 3, 4}
Output : 0
The array is already sorted
Input : arr[] = {3, 2, 1}
Output : 2
Step 1: L=1 and R=2 (0-based)
Step 2: L=2 and R=2
Resultant array [3, 3, 3]
Considering a sorted array, incrementing all elements of the array would still result in a sorted array.
So the idea is to traverse the elements of the array from right starting from the last index and keeping track of the minimum element. If at any point, the order of element is found to be increasing calculate the number of moves by subtracting the min element on right from current element.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int minMovesToSort( int arr[], int n)
{
int moves = 0;
int i, mn = arr[n - 1];
for (i = n - 2; i >= 0; i--) {
if (arr[i] > mn)
moves += arr[i] - mn;
mn = arr[i];
}
return moves;
}
int main()
{
int arr[] = { 3, 5, 2, 8, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << minMovesToSort(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int minMovesToSort( int arr[], int n)
{
int moves = 0 ;
int i, mn = arr[n - 1 ];
for (i = n - 2 ; i >= 0 ; i--) {
if (arr[i] > mn)
moves += arr[i] - mn;
mn = arr[i];
}
return moves;
}
public static void main (String[] args) {
int arr[] = { 3 , 5 , 2 , 8 , 4 };
int n = arr.length;
System.out.println( minMovesToSort(arr, n));
}
}
|
Python3
def minMovesToSort(arr, n) :
moves = 0
mn = arr[n - 1 ]
for i in range (n - 1 , - 1 , - 1 ) :
if (arr[i] > mn) :
moves + = arr[i] - mn
mn = arr[i]
return moves
if __name__ = = "__main__" :
arr = [ 3 , 5 , 2 , 8 , 4 ]
n = len (arr)
print (minMovesToSort(arr, n))
|
C#
using System;
class GFG
{
static int minMovesToSort( int []arr,
int n)
{
int moves = 0;
int i, mn = arr[n - 1];
for (i = n - 2; i >= 0; i--)
{
if (arr[i] > mn)
moves += arr[i] - mn;
mn = arr[i];
}
return moves;
}
static public void Main ()
{
int []arr = { 3, 5, 2, 8, 4 };
int n = arr.Length;
Console.WriteLine(minMovesToSort(arr, n));
}
}
|
PHP
<?php
function minMovesToSort( $arr , $n )
{
$moves = 0;
$mn = $arr [ $n - 1];
for ( $i = $n - 2; $i >= 0; $i --)
{
if ( $arr [ $i ] > $mn )
$moves += $arr [ $i ] - $mn ;
$mn = $arr [ $i ];
}
return $moves ;
}
$arr = array (3, 5, 2, 8, 4 );
$n = sizeof( $arr );
echo minMovesToSort( $arr , $n );
?>
|
Javascript
<script>
function minMovesToSort(arr, n)
{
var moves = 0;
var i, mn = arr[n - 1];
for (i = n - 2; i >= 0; i--)
{
if (arr[i] > mn)
moves += arr[i] - mn;
mn = arr[i];
}
return moves;
}
var arr = [ 3, 5, 2, 8, 4 ];
var n = arr.length;
document.write(minMovesToSort(arr, n));
</script>
|
Time Complexity: O(N)
Auxiliary space: O(1)
Last Updated :
22 Sep, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...