计算每个数组元素的最大子序列

给定一个由N 个唯一元素组成的数组arr[] ，任务是生成一个长度为N的数组B[] ，使得B[i]是子序列的数量 其中arr[i]是最大元素。

例子：

Input: arr[] = {2, 3, 1}
Output: {2, 4, 1}
Explanation: Subsequences in which arr[0] ( = 2) is maximum are {2}, {2, 1}.
Subsequences in which arr[1] ( = 3) is maximum are {3}, {1, 3, 2}, {2, 3}, {1, 3}.
Subsequence in which arr[2] ( = 1) is maximum is {1}.

Input: arr[] = {23, 34, 12, 7, 15, 31}
Output: {8, 32, 2, 1, 4, 16}

方法：通过观察元素arr[i]将出现为最大值的所有子序列将包含所有小于arr[i]的元素，可以解决该问题。因此，不同子序列的总数将为2 ^{(小于 arr[i] 的元素数)} 。请按照以下步骤解决问题：

1. 根据给定数组中存在的相应值对数组arr[]索引进行排序，并将该索引存储在数组索引 [] 中，其中arr[indices[i]] < arr[indices[i+1]] 。
2. 用1初始化一个整数subseq来存储可能的子序列的数量。
3. 使用变量i在[0, N-1]范围内使用指针迭代N次。
   1. B[indices[i]]是其中arr[indices[i]]最大的子序列的数量，即2 ⁱ ，因为将有i 个元素小于arr[indices[i]]。
   2. 将B[indices[i]]的答案存储为B[indices[i]] = subseq 。
   3. 通过将subseq乘以2 来更新它。
4. 打印数组B[]的元素作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
void merge(int* indices, int* a, int l,
           int mid, int r)
{
    int temp_ind[r - l + 1], j = mid + 1;
    int i = 0, temp_l = l, k;
    while (l <= mid && j <= r) {
 
        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];
 
        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }
 
    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];
 
    // Add remainging elements
    while (j <= r)
        temp_ind[i++] = indices[j++];
    for (k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}
 
// Recursive function to divide
// the array into to parts
void divide(int* indices, int* a, int l, int r)
{
    if (l >= r)
        return;
    int mid = l / 2 + r / 2;
 
    // Recursive call for elements before mid
    divide(indices, a, l, mid);
 
    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);
 
    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}
 
// Function to find the number of
// subsequences for each element
void noOfSubsequences(int arr[], int N)
{
    int indices[N], i;
    for (i = 0; i < N; i++)
        indices[i] = i;
 
    // Sorting the indices according
    // to array arr[]
    divide(indices, arr, 0, N - 1);
 
    // Array to store output nmbers
    int B[N];
 
    // Initialize subseq
    int subseq = 1;
    for (i = 0; i < N; i++) {
 
        // B[i] is 2^i
        B[indices[i]] = subseq;
 
        // Doubling the subsequences
        subseq *= 2;
    }
    // Print the final output, array B[]
    for (i = 0; i < N; i++)
        cout << B[i] << " ";
}
 
// Driver Code
int main()
{
 
    // Given array
    int arr[] = { 2, 3, 1 };
 
    // Given length
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    noOfSubsequences(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
static void merge(int[] indices, int[] a, int l,
                  int mid, int r)
{
    int []temp_ind = new int[r - l + 1];
    int j = mid + 1;
    int i = 0, temp_l = l, k;
     
    while (l <= mid && j <= r)
    {
         
        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];
 
        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }
 
    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];
 
    // Add remainging elements
    while (j <= r)
        temp_ind[i++] = indices[j++];
         
    for(k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}
 
// Recursive function to divide
// the array into to parts
static void divide(int[] indices, int[] a,
                   int l, int r)
{
    if (l >= r)
        return;
         
    int mid = l / 2 + r / 2;
 
    // Recursive call for elements before mid
    divide(indices, a, l, mid);
 
    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);
 
    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}
 
// Function to find the number of
// subsequences for each element
static void noOfSubsequences(int arr[], int N)
{
    int []indices = new int[N];
    int i;
     
    for(i = 0; i < N; i++)
        indices[i] = i;
 
    // Sorting the indices according
    // to array arr[]
    divide(indices, arr, 0, N - 1);
 
    // Array to store output nmbers
    int[] B = new int[N];
 
    // Initialize subseq
    int subseq = 1;
     
    for(i = 0; i < N; i++)
    {
         
        // B[i] is 2^i
        B[indices[i]] = subseq;
 
        // Doubling the subsequences
        subseq *= 2;
    }
     
    // Print the final output, array B[]
    for(i = 0; i < N; i++)
        System.out.print(B[i] + " ");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 2, 3, 1 };
 
    // Given length
    int N = arr.length;
 
    // Function call
    noOfSubsequences(arr, N);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
 
# Function to merge the subarrays
# arr[l .. m] and arr[m + 1, .. r]
# based on indices[]
def merge(indices, a, l, mid, r):
 
    temp_ind = [0] * (r - l + 1)
    j = mid + 1
    i = 0
    temp_l = l
     
    while (l <= mid and j <= r):
         
        # If a[indices[l]] is less than
        # a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]]):
            temp_ind[i] = indices[l]
            i += 1
            l += 1
 
        # Else add indices[j]
        else:
            temp_ind[i] = indices[j]
            i += 1
            j += 1
 
    # Add remaining elements
    while (l <= mid):
        temp_ind[i] = indices[l]
        i += 1
        l += 1
 
    # Add remainging elements
    while (j <= r):
        temp_ind[i] = indices[j]
        i += 1
        j += 1
         
    for k in range(i):
        indices[temp_l] = temp_ind[k]
        temp_l += 1
 
# Recursive function to divide
# the array into to parts
def divide(indices, a, l, r):
 
    if (l >= r):
        return
     
    mid = l // 2 + r // 2
 
    # Recursive call for elements
    # before mid
    divide(indices, a, l, mid)
 
    # Recursive call for elements
    # after mid
    divide(indices, a, mid + 1, r)
 
    # Merge the two sorted arrays
    merge(indices, a, l, mid, r)
 
# Function to find the number of
# subsequences for each element
def noOfSubsequences(arr, N):
 
    indices = N * [0]
    for i in range(N):
        indices[i] = i
 
    # Sorting the indices according
    # to array arr[]
    divide(indices, arr, 0, N - 1)
 
    # Array to store output nmbers
    B = [0] * N
 
    # Initialize subseq
    subseq = 1
    for i in range(N):
 
        # B[i] is 2^i
        B[indices[i]] = subseq
 
        # Doubling the subsequences
        subseq *= 2
 
    # Print the final output, array B[]
    for i in range(N):
        print(B[i], end = " ")
 
# Driver Code
if __name__ == "__main__":
 
    # Given array
    arr = [ 2, 3, 1 ]
 
    # Given length
    N = len(arr)
 
    # Function call
    noOfSubsequences(arr, N)
 
# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
static void merge(int[] indices, int[] a, int l,
                  int mid, int r)
{
    int []temp_ind = new int[r - l + 1];
    int j = mid + 1;
    int i = 0, temp_l = l, k;
     
    while (l <= mid && j <= r)
    {
         
        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];
 
        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }
 
    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];
 
    // Add remainging elements
    while (j <= r)
        temp_ind[i++] = indices[j++];
         
    for(k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}
 
// Recursive function to divide
// the array into to parts
static void divide(int[] indices, int[] a,
                   int l, int r)
{
    if (l >= r)
        return;
         
    int mid = l / 2 + r / 2;
 
    // Recursive call for elements before mid
    divide(indices, a, l, mid);
 
    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);
 
    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}
 
// Function to find the number of
// subsequences for each element
static void noOfSubsequences(int []arr, int N)
{
    int []indices = new int[N];
    int i;
     
    for(i = 0; i < N; i++)
        indices[i] = i;
 
    // Sorting the indices according
    // to array []arr
    divide(indices, arr, 0, N - 1);
 
    // Array to store output nmbers
    int[] B = new int[N];
 
    // Initialize subseq
    int subseq = 1;
     
    for(i = 0; i < N; i++)
    {
         
        // B[i] is 2^i
        B[indices[i]] = subseq;
 
        // Doubling the subsequences
        subseq *= 2;
    }
     
    // Print the readonly output, array []B
    for(i = 0; i < N; i++)
        Console.Write(B[i] + " ");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 2, 3, 1 };
 
    // Given length
    int N = arr.Length;
 
    // Function call
    noOfSubsequences(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// JavaScript program for the above approach
 
// Function to merge the subarrays
// arr[l .. m] and arr[m + 1, .. r]
// based on indices[]
function merge(indices, a, l, mid, r)
{
    let temp_ind = [];
    let j = mid + 1;
    let i = 0, temp_l = l, k;
      
    while (l <= mid && j <= r)
    {
          
        // If a[indices[l]] is less than
        // a[indices[j]], add indice[l] to temp
        if (a[indices[l]] < a[indices[j]])
            temp_ind[i++] = indices[l++];
  
        // Else add indices[j]
        else
            temp_ind[i++] = indices[j++];
    }
  
    // Add remaining elements
    while (l <= mid)
        temp_ind[i++] = indices[l++];
  
    // Add remainging elements
    while (j <= r)
        temp_ind[i++] = indices[j++];
          
    for(k = 0; k < i; k++)
        indices[temp_l++] = temp_ind[k];
}
  
// Recursive function to divide
// the array leto to parts
function divide(indices, a, l, r)
{
    if (l >= r)
        return;
          
    let mid = l / 2 + r / 2;
  
    // Recursive call for elements before mid
    divide(indices, a, l, mid);
  
    // Recursive call for elements after mid
    divide(indices, a, mid + 1, r);
  
    // Merge the two sorted arrays
    merge(indices, a, l, mid, r);
}
  
// Function to find the number of
// subsequences for each element
function noOfSubsequences(arr, N)
{
    let indices = [];
    let i;
      
    for(i = 0; i < N; i++)
        indices[i] = i;
  
    // Sorting the indices according
    // to array arr[]
    divide(indices, arr, 0, N - 1);
  
    // Array to store output nmbers
    let B = [];
  
    // Initialize subseq
    let subseq = 1;
      
    for(i = 0; i < N; i++)
    {
          
        // B[i] is 2^i
        B[indices[i]] = subseq;
  
        // Doubling the subsequences
        subseq *= 2;
    }
      
    // Print the final output, array B[]
    for(i = 0; i < N; i++)
        document.write(B[i] + " ");
}
 
// Driver code
 
// Given array
let arr = [ 2, 3, 1 ];
 
// Given length
let N = arr.length;
 
// Function call
noOfSubsequences(arr, N);
 
// This code is contributed by avijitmondal1998
 

输出：

2 4 1

时间复杂度： O(NlogN)，其中 N 是给定数组的长度。
辅助空间： O(N)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live