获得至少两个相等的 Array 元素所需的索引值的最小增量

给定一个由N 个整数组成的严格递减数组arr[] ，任务是找到使至少两个数组元素相等所需的最少操作次数，其中每个操作涉及将每个数组元素增加其索引值。

例子：

Input: arr[] = {6, 5, 1} 
Output: 1 
Explanation: 
{6 + 1, 5 + 2, 1 + 3} = {7, 7, 4} 
Input: arr[] = {12, 8, 4} 
Output: 4 
Explanation: 
Step 1 : {12 + 1, 8 + 2, 4 + 3} = {13, 10, 7} 
Step 2 : {13 + 1, 10 + 2, 7 + 3} = {14, 12, 10} 
Step 3 : {15, 14, 13} 
Step 4 : {16, 16, 16}

幼稚的方法：按照以下步骤解决问题：

• 检查数组是否已经至少有两个相等的元素。如果发现为真，则打印0 。
• 否则，通过将每个数组元素增加其索引值并增加count 来继续更新数组。检查数组是否有两个相等的元素。
• 一旦发现数组包含至少两个相等的元素，就打印计数。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to update every element
// adding to it its index value
void update(int arr[], int N)
{
    for (int i = 0; i < N; i++) {
        arr[i] += (i + 1);
    }
}
 
// Function to check if at least
// two elements are equal or not
bool check(int arr[], int N)
{
    bool f = 0;
    for (int i = 0; i < N; i++) {
 
        // Count the frequency of arr[i]
        int count = 0;
        for (int j = 0; j < N; j++) {
 
            if (arr[i] == arr[j]) {
                count++;
            }
        }
 
        if (count >= 2) {
            f = 1;
            break;
        }
    }
    if (f == 1)
        return true;
    else
        return false;
}
 
// Funtion to calculate the number
// of increment operations required
void incrementCount(int arr[], int N)
{
    // Stores the minimum number of steps
    int min = 0;
 
    while (check(arr, N) != true) {
        update(arr, N);
        min++;
    }
 
    cout << min;
}
 
// Driver Code
int main()
{
    int N = 3;
 
    int arr[N] = { 12, 8, 4 };
 
    incrementCount(arr, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to update every element
// adding to it its index value
static void update(int arr[], int N)
{
    for(int i = 0; i < N; i++)
    {
        arr[i] += (i + 1);
    }
}
 
// Function to check if at least
// two elements are equal or not
static boolean check(int arr[], int N)
{
    int f = 0;
    for(int i = 0; i < N; i++)
    {
         
        // Count the frequency of arr[i]
        int count = 0;
        for(int j = 0; j < N; j++)
        {
            if (arr[i] == arr[j])
            {
                count++;
            }
        }
 
        if (count >= 2)
        {
            f = 1;
            break;
        }
    }
    if (f == 1)
        return true;
    else
        return false;
}
 
// Funtion to calculate the number
// of increment operations required
static void incrementCount(int arr[], int N)
{
     
    // Stores the minimum number of steps
    int min = 0;
 
    while (check(arr, N) != true)
    {
        update(arr, N);
        min++;
    }
    System.out.println(min);
}
     
// Driver code
public static void main (String[] args)
{
    int N = 3;
    int arr[] = { 12, 8, 4 };
     
    incrementCount(arr, N);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
 
# Function to update every element
# adding to it its index value
def update(arr, N):
     
    for i in range(N):
        arr[i] += (i + 1);
 
# Function to check if at least
# two elements are equal or not
def check(arr, N):
     
    f = 0;
    for i in range(N):
 
        # Count the frequency of arr[i]
        count = 0;
         
        for j in range(N):
            if (arr[i] == arr[j]):
                count += 1;
 
        if (count >= 2):
            f = 1;
            break;
 
    if (f == 1):
        return True;
    else:
        return False;
 
# Funtion to calculate the number
# of increment operations required
def incrementCount(arr, N):
     
    # Stores the minimum number of steps
    min = 0;
 
    while (check(arr, N) != True):
        update(arr, N);
        min += 1;
 
    print(min);
 
# Driver code
if __name__ == '__main__':
     
    N = 3;
    arr = [ 12, 8, 4 ];
 
    incrementCount(arr, N);
 
# This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to update every element
// adding to it its index value
static void update(int []arr, int N)
{
    for(int i = 0; i < N; i++)
    {
        arr[i] += (i + 1);
    }
}
 
// Function to check if at least
// two elements are equal or not
static bool check(int []arr, int N)
{
    int f = 0;
    for(int i = 0; i < N; i++)
    {
         
        // Count the frequency of arr[i]
        int count = 0;
        for(int j = 0; j < N; j++)
        {
            if (arr[i] == arr[j])
            {
                count++;
            }
        }
 
        if (count >= 2)
        {
            f = 1;
            break;
        }
    }
    if (f == 1)
        return true;
    else
        return false;
}
 
// Funtion to calculate the number
// of increment operations required
static void incrementCount(int []arr, int N)
{
     
    // Stores the minimum number of steps
    int min = 0;
 
    while (check(arr, N) != true)
    {
        update(arr, N);
        min++;
    }
    Console.WriteLine(min);
}
     
// Driver code
public static void Main(String[] args)
{
    int N = 3;
    int []arr = { 12, 8, 4 };
     
    incrementCount(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

     
 
// Java Script program to implement
// the above approach
 
 
// Function to update every element
// adding to it its index value
function update(arr,N)
{
    for(let i = 0; i < N; i++)
    {
        arr[i] += (i + 1);
    }
}
 
// Function to check if at least
// two elements are equal or not
function check(arr,N)
{
    let f = 0;
    for(let i = 0; i < N; i++)
    {
         
        // Count the frequency of arr[i]
        let count = 0;
        for(let j = 0; j < N; j++)
        {
            if (arr[i] == arr[j])
            {
                count++;
            }
        }
 
        if (count >= 2)
        {
            f = 1;
            break;
        }
    }
    if (f == 1)
        return true;
    else
        return false;
}
 
// Funtion to calculate the number
// of increment operations required
function incrementCount(arr,N)
{
     
    // Stores the minimum number of steps
    let min = 0;
 
    while (check(arr, N) != true)
    {
        update(arr, N);
        min++;
    }
    document.write(min);
}
     
// Driver code
 
    let N = 3;
    let arr= [12, 8, 4 ];
     
    incrementCount(arr, N);
 
//contributed by 171fa07058

输出：

4

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：通过观察给定的操作，任何两个相邻元素之间的差异随着数组的减少而减少1，可以优化上述方法。因此，所需的最少操作次数等于任意两个相邻元素之间的最小差值。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate the minimum
// number of steps required
void incrementCount(int arr[], int N)
{
    // Stores minimum difference
    int mini = arr[0] - arr[1];
 
    for (int i = 2; i < N; i++) {
 
        mini
            = min(mini, arr[i - 1] - arr[i]);
    }
 
    cout << mini;
}
 
// Driver Code
int main()
{
    int N = 3;
    int arr[N] = { 12, 8, 4 };
    incrementCount(arr, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to calculate the minimum
// number of steps required
static void incrementCount(int arr[], int N)
{
     
    // Stores minimum difference
    int mini = arr[0] - arr[1];
 
    for(int i = 2; i < N; i++)
    {
        mini = Math.min(mini,
                        arr[i - 1] - arr[i]);
    }
    System.out.println(mini);
}   
  
// Driver code
public static void main (String[] args)
{
    int N = 3;
    int arr[] = { 12, 8, 4 };
     
    incrementCount(arr, N);
}
}
 
// This code is contributed by offbeat

蟒蛇3

# Python3 program to implement
# the above approach
 
# Function to calculate the minimum
# number of steps required
def incrementCount(arr, N):
 
    # Stores minimum difference
    mini = arr[0] - arr[1]
 
    for i in range(2, N):
        mini = min(mini,
                   arr[i - 1] - arr[i])
 
    print(mini)
 
# Driver Code
N = 3
arr = [ 12, 8, 4 ]
 
# Function call
incrementCount(arr, N)
 
# This code is contributed by Shivam Singh

C＃

// C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to calculate the minimum
// number of steps required
static void incrementCount(int []arr, int N)
{
     
    // Stores minimum difference
    int mini = arr[0] - arr[1];
 
    for(int i = 2; i < N; i++)
    {
        mini = Math.Min(mini,
                        arr[i - 1] - arr[i]);
    }
    Console.WriteLine(mini);
}    
 
// Driver code
public static void Main(String[] args)
{
    int N = 3;
    int []arr = { 12, 8, 4 };
     
    incrementCount(arr, N);
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript program for the above approach
 
// Function to calculate the minimum
// number of steps required
function incrementCount(arr, N)
{
       
    // Stores minimum difference
    let mini = arr[0] - arr[1];
   
    for(let i = 2; i < N; i++)
    {
        mini = Math.min(mini,
                        arr[i - 1] - arr[i]);
    }
    document.write(mini);
}
     
// Driver Code
 
    let N = 3;
    let arr = [ 12, 8, 4 ];
       
    incrementCount(arr, N);
      

输出：

4

时间复杂度： O(N)
辅助空间： O(1)

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