最小增量操作以使阵列按升序排列

给定大小为N和X的数组。找到使数组按升序排列所需的最小移动。在每一步中，可以将X添加到数组中的任何元素。

例子：

Input : a = { 1, 3, 3, 2 }, X = 2
Output : 3
Explanation : Modified array is { 1, 3, 5, 6 }

Input : a = { 3, 5, 6 }, X = 5
Output : 0

观察：

Let’s take two numbers a and b. a >= b and convert this into a < b by adding some number X. 
so, a < b + k*X 
( a – b ) / x < k
so, the minimum possible value of k is ( a – b ) / x + 1. 
 

方法：

Iterate over the given array and take two numbers when a[i] >= a[i-1] 
and apply above observation.

下面是上述方法的实现：

C++

// C++ program to find minimum moves required
// to make the array in increasing order
#include 
using namespace std;
 
// function to find minimum moves required
// to make the array in increasing order
int MinimumMoves(int a[], int n, int x)
{
    // to store answer
    int ans = 0;
 
    // iterate over an array
    for (int i = 1; i < n; i++) {
 
        // non- increasing order
        if (a[i] <= a[i - 1]) {
            int p = (a[i - 1] - a[i]) / x + 1;
 
            // add moves to answer
            ans += p;
 
            // increase the element
            a[i] += p * x;
        }
    }
 
    // return required answer
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 3, 2 };
    int x = 2;
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << MinimumMoves(arr, n, x);
 
    return 0;
}

Java

// Java program to find minimum moves required
// to make the array in increasing order
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG{
// function to find minimum moves required
// to make the array in increasing order
static int MinimumMoves(int a[], int n, int x)
{
    // to store answer
    int ans = 0;
  
    // iterate over an array
    for (int i = 1; i < n; i++) {
  
        // non- increasing order
        if (a[i] <= a[i - 1]) {
            int p = (a[i - 1] - a[i]) / x + 1;
  
            // add moves to answer
            ans += p;
  
            // increase the element
            a[i] += p * x;
        }
    }
  
    // return required answer
    return ans;
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 3, 3, 2 };
    int x = 2;
    int n = arr.length;
  
    System.out.println(MinimumMoves(arr, n, x));
  
}
}

Python3

# Python3 program to find minimum
# moves required to make the array
# in increasing order
 
# function to find minimum moves required
# to make the array in increasing order
def MinimumMoves(a, n, x) :
 
    # to store answer
    ans = 0
 
    # iterate over an array
    for i in range(1, n) :
 
        # non- increasing order
        if a[i] <= a[i - 1] :
 
            p = (a[i - 1] - a[i]) // x + 1
 
            # add moves to answer
            ans += p
 
            # increase the element
            a[i] += p * x
 
    # return required answer
    return ans
         
# Driver code    
if __name__ == "__main__" :
 
    arr = [1, 3, 3, 2]
    x = 2
    n = len(arr)
 
    print(MinimumMoves(arr, n, x))
 
 
# This code is contributed by ANKITRAI1

C#

// C# program to find minimum moves required
// to make the array in increasing order
using System;
 
class GFG {
     
// function to find minimum moves required
// to make the array in increasing order
static int MinimumMoves(int[] a, int n, int x)
{
     
    // to store answer
    int ans = 0;
     
    // iterate over an array
    for (int i = 1; i < n; i++) {
     
        // non- increasing order
        if (a[i] <= a[i - 1]) {
             
            int p = (a[i - 1] - a[i]) / x + 1;
     
            // add moves to answer
            ans += p;
     
            // increase the element
            a[i] += p * x;
        }
    }
     
    // return required answer
    return ans;
}
     
// Driver code
public static void Main()
{
     
    int[] arr = {1, 3, 3, 2};
    int x = 2;
    int n = arr.Length;
     
    Console.Write(MinimumMoves(arr, n, x));
     
}
}
 
// This code is contributed by ChitraNayal

Javascript

 
// Javascript program to find minimum
// moves required to make the array in
// increasing order   
 
// Function to find minimum moves required
// to make the array in increasing order
function MinimumMoves(a, n, x)
{
     
    // To store answer
    var ans = 0;
 
    // Tterate over an array
    for(i = 1; i < n; i++)
    {
         
        // Non- increasing order
        if (a[i] <= a[i - 1])
        {
            var p = parseInt((a[i - 1] -
                              a[i]) / x + 1);
 
            // Add moves to answer
            ans += p;
 
            // Increase the element
            a[i] += p * x;
        }
    }
 
    // Return required answer
    return ans;
}
 
// Driver code
var arr = [ 1, 3, 3, 2 ];
var x = 2;
var n = arr.length;
 
document.write(MinimumMoves(arr, n, x));
 
// This code is contributed by aashish1995
 

输出：

3

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