通过将每个元素减少其下一个较小的元素来修改给定的数组

给定一个长度为N的数组arr[] ，如果可能的话，任务是通过将给定数组的每个元素替换为其下一个较小元素来修改给定数组。将修改后的数组打印为所需答案。

例子：

Input: arr[] = {8, 4, 6, 2, 3}
Output: 4 2 4 2 3
Explanation: The operations can be performed as follows:

• For arr[0], arr[1] is the next smaller element.
• For arr[1], arr[3] is the next smaller element.
• For arr[2], arr[3] is the next smaller element.
• For arr[3], no smaller element present after it.
• For arr[4], no smaller element present after it.

Input: arr[] = {1, 2, 3, 4, 5}
Output: 1 2 3 4 5

朴素的方法：最简单的方法是遍历数组，对于每个元素，遍历它后面的剩余元素并检查是否存在任何较小的元素。如果找到，则将该元素减少获得的第一个较小的元素。

时间复杂度： O(N ² )
辅助空间： O(N)

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to print the final array
// after reducing each array element
// by its next smaller element
void printFinalPrices(vector& arr)
{
    // Stores the resultant array
    vector ans;
 
    // Traverse the array
    for (int i = 0; i < arr.size(); i++) {
        int flag = 1;
        for (int j = i + 1; j < arr.size(); j++) {
 
            // If a smaller element is found
            if (arr[j] <= arr[i]) {
 
                // Reduce current element by
                // next smaller element
                ans.push_back(arr[i] - arr[j]);
                flag = 0;
                break;
            }
        }
 
        // If no smaller element is found
        if (flag == 1)
            ans.push_back(arr[i]);
    }
 
    // Print the answer
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i] << " ";
}
 
// Driver Code
int main()
{
    // Given array
    vector arr = { 8, 4, 6, 2, 3 };
 
    // Function Call
    printFinalPrices(arr);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to print the final array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
     
    // Stores the resultant array
    ArrayList ans = new ArrayList();
     
    // Traverse the array
    for(int i = 0; i < arr.length; i++)
    {
        int flag = 1;
        for(int j = i + 1; j < arr.length; j++)
        {
             
            // If a smaller element is found
            if (arr[j] <= arr[i])
            {
                 
                // Reduce current element by
                // next smaller element
                ans.add(arr[i] - arr[j]);
                flag = 0;
                break;
            }
        }
    
        // If no smaller element is found
        if (flag == 1)
            ans.add(arr[i]);
    }
    
    // Print the answer
    for(int i = 0; i < ans.size(); i++)
        System.out.print(ans.get(i) + " ");
} 
  
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int[] arr = { 8, 4, 6, 2, 3 };
   
    // Function Call
    printFinalPrices(arr);
}
}
 
// This code is contributed by code_hunt

Python3

# Python3 program for the above approach
 
# Function to print the final array
# after reducing each array element
# by its next smaller element
def printFinalarr(arr):
 
  # Stores resultant array
    ans = []
 
    # Traverse the given array
    for i in range(len(arr)):
        flag = 1
        for j in range(i + 1, len(arr)):
             
            # If a smaller element is found
            if arr[j] <= arr[i]:
 
                # Reduce current element by
                # next smaller element
                ans.append(arr[i] - arr[j])
                flag = 0
                break
        if flag:
 
            # If no smaller element is found
            ans.append(arr[i])
 
    # Print the final array
    for k in range(len(ans)):
        print(ans[k], end =' ')
 
 
# Driver Code
if __name__ == '__main__':
 
  # Given array
    arr = [8, 4, 6, 2, 3]
 
  # Function call
    printFinalarr(arr)

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to print the final array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
     
    // Stores the resultant array
    List ans = new List();
   
    // Traverse the array
    for(int i = 0; i < arr.Length; i++)
    {
        int flag = 1;
        for(int j = i + 1; j < arr.Length; j++)
        {
             
            // If a smaller element is found
            if (arr[j] <= arr[i])
            {
                 
                // Reduce current element by
                // next smaller element
                ans.Add(arr[i] - arr[j]);
                flag = 0;
                break;
            }
        }
   
        // If no smaller element is found
        if (flag == 1)
            ans.Add(arr[i]);
    }
   
    // Print the answer
    for(int i = 0; i < ans.Count; i++)
        Console.Write(ans[i] + " ");
} 
 
// Driver code
static void Main()
{
     
    // Given array
    int[] arr = { 8, 4, 6, 2, 3 };
  
    // Function Call
    printFinalPrices(arr);
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

// Js program for the above approach
// Function to print the final array
// after reducing each array element
// by its next smaller element
function printFinalPrices( arr)
{
    // Stores the resultant array
    let ans = [];
 
    // Traverse the array
    for (let i = 0; i < arr.length; i++) {
        let flag = 1;
        for (let j = i + 1; j < arr.length; j++) {
 
            // If a smaller element is found
            if (arr[j] <= arr[i]) {
 
                // Reduce current element by
                // next smaller element
                ans.push(arr[i] - arr[j]);
                flag = 0;
                break;
            }
        }
 
        // If no smaller element is found
        if (flag == 1)
            ans.push(arr[i]);
    }
 
    // Print the answer
    for (let i = 0; i < ans.length; i++)
        document.write(ans[i], " ");
}
 
// Driver Code
// Given array
    let arr = [ 8, 4, 6, 2, 3 ];
 
    // Function Call
    printFinalPrices(arr);
 

输出：

4 2 4 2 3

Efficient Approach：优化上述方式，思路是使用Stack数据结构。请按照以下步骤解决问题：

1. 初始化堆栈 和一个大小为N的数组ans[] ，用于存储结果数组。
2. 在索引i = N – 1 到 0 上遍历给定数组。
3. 如果堆栈为空，则将当前元素arr[i]压入堆栈顶部。
4. 否则，如果当前元素大于栈顶元素，则将其压入栈中，然后从栈中移除元素，直到栈为空或找到小于等于arr[i]的元素.之后，如果栈不为空，则设置ans[i] = arr[i] – 栈顶元素，然后将其从栈中移除。
5. 否则，从堆栈中移除顶部元素并将ans[i]设置为等于堆栈中的顶部元素，然后将其从堆栈中移除。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the final array
// after reducing each array element
// by its next smaller element
void printFinalPrices(vector& arr)
{
 
    // Initialize stack
    stack minStk;
 
    // Array size
    int n = arr.size();
 
    // To store the corresponding element
    vector reduce(n, 0);
    for (int i = n - 1; i >= 0; i--) {
 
        // If stack is not empty
        if (!minStk.empty()) {
 
            // If top element is smaller
            // than the current element
            if (minStk.top() <= arr[i]) {
                reduce[i] = minStk.top();
            }
            else {
 
                // Keep popping until stack is empty
                // or top element is greater than
                // the current element
                while (!minStk.empty()
                       && (minStk.top() > arr[i])) {
                    minStk.pop();
                }
 
                // If stack is not epty
                if (!minStk.empty()) {
                    reduce[i] = minStk.top();
                }
            }
        }
 
        // Push current element
        minStk.push(arr[i]);
    }
 
    // Print the final array
    for (int i = 0; i < n; i++)
        cout << arr[i] - reduce[i] << " ";
}
 
// Driver Code
int main()
{
 
    // Given array
    vector arr = { 8, 4, 6, 2, 3 };
 
    // Function call
    printFinalPrices(arr);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to print the final array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
     
    // Initialize stack
    Stack minStk = new Stack<>();
     
    // Array size
    int n = arr.length;
 
    // To store the corresponding element
    int[] reduce = new int[n];
    for(int i = n - 1; i >= 0; i--)
    {
         
        // If stack is not empty
        if (!minStk.isEmpty())
        {
             
            // If top element is smaller
            // than the current element
            if (minStk.peek() <= arr[i])
            {
                reduce[i] = minStk.peek();
            }
            else
            {
                 
                // Keep popping until stack is empty
                // or top element is greater than
                // the current element
                while (!minStk.isEmpty() &&
                       (minStk.peek() > arr[i]))
                {
                    minStk.pop();
                }
 
                // If stack is not epty
                if (!minStk.isEmpty())
                {
                    reduce[i] = minStk.peek();
                }
            }
        }
 
        // Push current element
        minStk.add(arr[i]);
    }
 
    // Print the final array
    for(int i = 0; i < n; i++)
        System.out.print(arr[i] - reduce[i] + " ");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int[] arr = { 8, 4, 6, 2, 3 };
 
    // Function call
    printFinalPrices(arr);
}
}
 
// This code is contributed by Rajput-Ji

蟒蛇3

# Python3 program for the above approach
 
# Function to print the final array
# after reducing each array element
# by its next smaller element
def printFinalPrices(arr):
 
  # Initialize stack
    minStk = []
 
    # To store the corresponding element
    reduce = [0] * len(arr)
    for i in range(len(arr) - 1, -1, -1):
 
       # If stack is not empty
        if minStk:
 
            # If top element is smaller
            # than the current element
            if minStk[-1] <= arr[i]:
                reduce[i] = minStk[-1]
            else:
 
              # Keep popping until stack is empty
                # or top element is greater than
                # the current element
                while minStk and minStk[-1] > arr[i]:
                    minStk.pop()
 
                if minStk:
 
                  # Corresponding elements
                    reduce[i] = minStk[-1]
 
        # Push current element
        minStk.append(arr[i])
 
    # Final array
    for i in range(len(arr)):
        print(arr[i] - reduce[i], end =' ')
 
 
# Driver Code
if __name__ == '__main__':
 
  # Given array
    arr = [8, 4, 6, 2, 3]
 
   # Function Call
    printFinalPrices(arr)

C＃

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to print the readonly array
// after reducing each array element
// by its next smaller element
static void printFinalPrices(int[] arr)
{
     
    // Initialize stack
    Stack minStk = new Stack();
     
    // Array size
    int n = arr.Length;
 
    // To store the corresponding element
    int[] reduce = new int[n];
    for(int i = n - 1; i >= 0; i--)
    {
         
        // If stack is not empty
        if (minStk.Count != 0)
        {
             
            // If top element is smaller
            // than the current element
            if (minStk.Peek() <= arr[i])
            {
                reduce[i] = minStk.Peek();
            }
            else
            {
                 
                // Keep popping until stack is empty
                // or top element is greater than
                // the current element
                while (minStk.Count != 0 &&
                       (minStk.Peek() > arr[i]))
                {
                    minStk.Pop();
                }
 
                // If stack is not epty
                if (minStk.Count != 0)
                {
                    reduce[i] = minStk.Peek();
                }
            }
        }
 
        // Push current element
        minStk.Push(arr[i]);
    }
 
    // Print the readonly array
    for(int i = 0; i < n; i++)
        Console.Write(arr[i] - reduce[i] + " ");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int[] arr = { 8, 4, 6, 2, 3 };
 
    // Function call
    printFinalPrices(arr);
}
}
 
// This code contributed by shikhasingrajput

Javascript

 
// javascript program for the above approach
 
// Function to print the final array
// after reducing each array element
// by its next smaller element
function printFinalPrices(arr)
{
 
    // Initialize stack
    var minStk = []
 
    // Array size
    var n = arr.length;
    var i;
 
    // To store the corresponding element
    var reduce = Array(n).fill(0);
    for (i = n - 1; i >= 0; i--) {
 
        // If stack is not empty
        if (minStk.length>0) {
 
            // If top element is smaller
            // than the current element
            if (minStk[minStk.length-1] <= arr[i]) {
                reduce[i] = minStk[minStk.length-1];
            }
            else {
 
                // Keep popping until stack is empty
                // or top element is greater than
                // the current element
                while (minStk.length>0
                       && (minStk[minStk.length-1] > arr[i])) {
                    minStk.pop();
                }
 
                // If stack is not epty
                if (minStk.length>0) {
                    reduce[i] = minStk[minStk.length-1];
                }
            }
        }
 
        // Push current element
        minStk.push(arr[i]);
    }
 
    // Print the final array
    for (i = 0; i < n; i++)
        document.write(arr[i] - reduce[i] + " ");
}
 
// Driver Code
 
    // Given array
    var arr = [8, 4, 6, 2, 3];
 
    // Function call
    printFinalPrices(arr);
 
// This code is contributed by ipg2016107.

输出：

4 2 4 2 3

时间复杂度： O(N)
辅助空间： O(N)

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