最小化所需的奖赏数量，以便较小的值在相邻对中获得更少的奖赏

给定长度为N的数组arr [] ，任务是找到所需的最少奖赏数量，使得如果两个元素相邻，则价值较小的元素与其价值较大的相邻元素相比，奖赏数量更少。
注意：每个元素将至少获得一个奖项。

例子：

Input: arr[] = {1, 2, 2, 3} 
Output: 6 
Explanation: 
Element at index {0} will get {1} prize. 
Element at index {1} will get {2} prizes. 
Element at index {2} will get {1} prizes. 
Element at index {3} will get {2} prizes. 
So, the total number of prizes required to satisfy 
the above conditions are 6

Input: arr[] = {3, 2, 2, 1} 
Output: 6 
Explanation: 
Element at index {0} will get {2} prize. 
Element at index {1} will get {1} prizes. 
Element at index {2} will get {2} prizes. 
Element at index {3} will get {1} prizes. 
So, the total number of prizes required to satisfy 
the above conditions are 6

天真的方法：遍历数组的元素，并为数组的每个元素在元素的左侧找到连续的较小元素，并在该索引的右侧找到连续的较小元素。

Prize at index i = max(Consecutive smaller elements at left, Consecutive smaller elements at right, 1)

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
 
#include 
 
using namespace std;
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
int findMinPrizes(int arr[], int n)
{
    int totalPrizes = 0, j, x, y;
 
    // Loop to iterate over every
    // elements of the array
    for (int i = 0; i < n; i++) {
        x = 1;
        j = i;
 
        // Loop to find the consecutive
        // smaller elements at left
        while (j > 0 && arr[j] > arr[j - 1]) {
            x++;
            j--;
        }
        j = i;
        y = 1;
 
        // Loop to find the consecutive
        // smaller elements at right
        while (j < n - 1 && arr[j] > arr[j + 1]) {
            y++;
            j++;
        }
 
        totalPrizes += max({ x, y });
    }
    cout << totalPrizes << endl;
 
    return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findMinPrizes(arr, n);
}

Java

// Java implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
import java.util.*;
class GFG{
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
static int findMinPrizes(int arr[], int n)
{
    int totalPrizes = 0, j, x, y;
 
    // Loop to iterate over every
    // elements of the array
    for (int i = 0; i < n; i++)
    {
        x = 1;
        j = i;
 
        // Loop to find the consecutive
        // smaller elements at left
        while (j > 0 && arr[j] > arr[j - 1])
        {
            x++;
            j--;
        }
        j = i;
        y = 1;
 
        // Loop to find the consecutive
        // smaller elements at right
        while (j < n - 1 && arr[j] > arr[j + 1])
        {
            y++;
            j++;
        }
 
        totalPrizes += Math.max(x, y );
    }
    System.out.print(totalPrizes + "\n");
 
    return 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 3 };
    int n = arr.length;
 
    findMinPrizes(arr, n);
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 implementation to find the
# minimum prizes required such
# that adjacent smaller elements
# gets less number of prizes
 
# Function to find the minimum
# number of required such that
# adjacent smaller elements gets
# less number of prizes
def findMinPrizes(arr, n):
     
    totalPrizes = 0
     
    # Loop to iterate over every
    # elements of the array
    for i in range(n):
        x = 1
        j = i
         
        # Loop to find the consecutive
        # smaller elements at left
        while (j > 0 and arr[j] > arr[j - 1]):
            x += 1
            j -= 1
         
        j = i
        y = 1
         
        # Loop to find the consecutive
        # smaller elements at right
        while (j < n - 1 and arr[j] >
                             arr[j + 1]):
            y += 1
            j += 1
             
        totalPrizes += max(x, y)
         
    print(totalPrizes)
 
# Driver code
arr = [ 1, 2, 2, 3 ]
n = len(arr)
 
findMinPrizes(arr, n)
 
# This code is contributed by stutipathak31jan

C#

// C# implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
using System;
 
class GFG{
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
static int findMinPrizes(int []arr, int n)
{
    int totalPrizes = 0, j, x, y;
 
    // Loop to iterate over every
    // elements of the array
    for(int i = 0; i < n; i++)
    {
        x = 1;
        j = i;
 
        // Loop to find the consecutive
        // smaller elements at left
        while (j > 0 && arr[j] > arr[j - 1])
        {
            x++;
            j--;
        }
        j = i;
        y = 1;
 
        // Loop to find the consecutive
        // smaller elements at right
        while (j < n - 1 &&
          arr[j] > arr[j + 1])
        {
            y++;
            j++;
        }
        totalPrizes += Math.Max(x, y);
    }
    Console.Write(totalPrizes + "\n");
 
    return 0;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 2, 3 };
    int n = arr.Length;
 
    findMinPrizes(arr, n);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
    // Javascript implementation to find the
    // minimum prizes required such
    // that adjacent smaller elements
    // gets less number of prizes 
     
    // Function to find the minimum
    // number of required such that
    // adjacent smaller elements gets
    // less number of prizes
    function findMinPrizes(arr, n)
    {
        let totalPrizes = 0, j, x, y;
 
        // Loop to iterate over every
        // elements of the array
        for (let i = 0; i < n; i++) {
            x = 1;
            j = i;
 
            // Loop to find the consecutive
            // smaller elements at left
            while (j > 0 && arr[j] > arr[j - 1]) {
                x++;
                j--;
            }
            j = i;
            y = 1;
 
            // Loop to find the consecutive
            // smaller elements at right
            while (j < n - 1 && arr[j] > arr[j + 1]) {
                y++;
                j++;
            }
 
            totalPrizes += Math.max( x, y );
        }
        document.write(totalPrizes);
 
        return 0;
    }
     
    let arr = [ 1, 2, 2, 3 ];
    let n = arr.length;
    findMinPrizes(arr, n);
     
    // This code is contributed by divyesh072019.

输出：

6

性能分析：

• 时间复杂度： O(N ² )
• 辅助空间： O(1)

高效方法：想法是为数组的每个元素预先计算左右连续的较小元素的数量。这意味着，如果左侧的元素较小，则该元素左侧的所有较小元素也将小于当前元素。 IE

if (arr[i-1] < arr[i])
    smallerLeft[i] = smallerLeft[i-1] + 1

类似地，可以使用以下事实来计算连续的较小元素：如果右侧的元素大于当前元素，那么右侧的连续的较大元素也将大于当前元素。 IE

if (arr[i] < arr[i+1])
    smallerRight[i] = smallerRight[i+1] + 1

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
 
#include 
 
using namespace std;
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
int minPrizes(int arr[], int n)
{
    int dpLeft[n];
 
    dpLeft[0] = 1;
 
    // Loop to compute the smaller
    // elements at the left
    for (int i = 1; i < n; i++) {
 
        if (arr[i] > arr[i - 1]) {
 
            dpLeft[i] = dpLeft[i - 1] + 1;
        }
        else {
 
            dpLeft[i] = 1;
        }
    }
 
    int dpRight[n];
 
    dpRight[n - 1] = 1;
 
    // Loop to find the smaller
    // elements at the right
    for (int i = n - 2; i >= 0; i--) {
 
        if (arr[i] > arr[i + 1]) {
 
            dpRight[i] = dpRight[i + 1] + 1;
        }
        else {
 
            dpRight[i] = 1;
        }
    }
 
    int totalPrizes = 0;
 
    // Loop to find the minimum
    // prizes required
    for (int i = 0; i < n; i++) {
 
        totalPrizes += max(dpLeft[i],
                           dpRight[i]);
    }
    cout << totalPrizes << endl;
 
    return 0;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    minPrizes(arr, n);
 
    return 0;
}

Java

// Java implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
import java.util.*;
 
class GFG{
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
static int minPrizes(int arr[], int n)
{
    int []dpLeft = new int[n];
 
    dpLeft[0] = 1;
 
    // Loop to compute the smaller
    // elements at the left
    for(int i = 1; i < n; i++)
    {
        if (arr[i] > arr[i - 1])
        {
            dpLeft[i] = dpLeft[i - 1] + 1;
        }
        else
        {
            dpLeft[i] = 1;
        }
    }
 
    int []dpRight = new int[n];
 
    dpRight[n - 1] = 1;
 
    // Loop to find the smaller
    // elements at the right
    for(int i = n - 2; i >= 0; i--)
    {
        if (arr[i] > arr[i + 1])
        {
            dpRight[i] = dpRight[i + 1] + 1;
        }
        else
        {
            dpRight[i] = 1;
        }
    }
 
    int totalPrizes = 0;
 
    // Loop to find the minimum
    // prizes required
    for(int i = 0; i < n; i++)
    {
        totalPrizes += Math.max(dpLeft[i],
                               dpRight[i]);
    }
    System.out.print(totalPrizes + "\n");
    return 0;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 3 };
    int n = arr.length;
 
    minPrizes(arr, n);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 implementation to find the
# minimum prizes required such
# that adjacent smaller elements
# gets less number of prizes
 
# Function to find the minimum
# number of required such that
# adjacent smaller elements gets
# less number of prizes
def minPrizes(arr, n):
     
    dpLeft = [0] * n
    dpLeft[0] = 1
     
    # Loop to compute the smaller
    # elements at the left
    for i in range(n):
        if arr[i] > arr[i - 1]:
            dpLeft[i] = dpLeft[i - 1] + 1
             
        else:
            dpLeft[i] = 1
         
    dpRight = [0] * n
    dpRight[-1] = 1
     
    # Loop to find the smaller
    # elements at the right
    for i in range(n - 2, -1, -1):
        if arr[i] > arr[i + 1]:
            dpRight[i] = dpRight[i + 1] + 1
             
        else:
            dpRight[i] = 1
     
    totalPrizes = 0
     
    # Loop to find the minimum
    # prizes required
    for i in range(n):
        totalPrizes += max(dpLeft[i],
                           dpRight[i])
         
    print(totalPrizes)
 
# Driver code
arr = [ 1, 2, 2, 3 ]
n = len(arr)
     
minPrizes(arr, n)
 
# This code is contributed by stutipathak31jan

C＃

// C# implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
using System;
 
class GFG{
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
static int minPrizes(int []arr, int n)
{
    int []dpLeft = new int[n];
 
    dpLeft[0] = 1;
 
    // Loop to compute the smaller
    // elements at the left
    for(int i = 1; i < n; i++)
    {
        if (arr[i] > arr[i - 1])
        {
            dpLeft[i] = dpLeft[i - 1] + 1;
        }
        else
        {
            dpLeft[i] = 1;
        }
    }
 
    int []dpRight = new int[n];
 
    dpRight[n - 1] = 1;
 
    // Loop to find the smaller
    // elements at the right
    for(int i = n - 2; i >= 0; i--)
    {
        if (arr[i] > arr[i + 1])
        {
            dpRight[i] = dpRight[i + 1] + 1;
        }
        else
        {
            dpRight[i] = 1;
        }
    }
 
    int totalPrizes = 0;
 
    // Loop to find the minimum
    // prizes required
    for(int i = 0; i < n; i++)
    {
        totalPrizes += Math.Max(dpLeft[i],
                               dpRight[i]);
    }
    Console.Write(totalPrizes + "\n");
    return 0;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 2, 3 };
    int n = arr.Length;
 
    minPrizes(arr, n);
}
}
 
// This code is contributed by Rajput-Ji

Java脚本

 
// Javascript implementation to find the
// minimum prizes required such
// that adjacent smaller elements
// gets less number of prizes
 
// Function to find the minimum
// number of required such that
// adjacent smaller elements gets
// less number of prizes
function minPrizes(arr, n)
{
    let dpLeft = new Array(n);
 
    dpLeft[0] = 1;
 
    // Loop to compute the smaller
    // elements at the left
    for(let i = 1; i < n; i++)
    {
        if (arr[i] > arr[i - 1])
        {
            dpLeft[i] = dpLeft[i - 1] + 1;
        }
        else
        {
            dpLeft[i] = 1;
        }
    }
 
    let dpRight = new Array(n);
 
    dpRight[n - 1] = 1;
 
    // Loop to find the smaller
    // elements at the right
    for(let i = n - 2; i >= 0; i--)
    {
        if (arr[i] > arr[i + 1])
        {
            dpRight[i] = dpRight[i + 1] + 1;
        }
        else
        {
            dpRight[i] = 1;
        }
    }
 
    let totalPrizes = 0;
 
    // Loop to find the minimum
    // prizes required
    for(let i = 0; i < n; i++)
    {
        totalPrizes += Math.max(dpLeft[i],
                               dpRight[i]);
    }
    document.write(totalPrizes + "</br>");
    return 0;
}
 
// Driver code
let arr = [ 1, 2, 2, 3 ];
let n = arr.length;
 
minPrizes(arr, n);
 
// This code is contributed by suresh07
 

输出：

6

性能分析：

• 时间复杂度： O(N)
• 辅助空间： O(N)