生成 1 到 N 的排列，每个元素的前缀最小值之和为 Y

给定两个整数N , Y ，生成长度为N的排列，使得该排列的所有前缀最小值之和为Y 。

例子：

Input: N = 5, Y = 10
Output: 5 2 1 4 3
Explanation:  Array of prefix minimum for [5, 2, 1, 4, 3] is [5, 2, 1, 1, 1]. Sum of this array of prefix minimum is 10 (= Y).

Input: N = 5, Y = 5
Output: 1 2 3 4 5
Explanation:  Array of prefix minimum for [1, 2, 3, 4, 5] is [1, 1, 1, 1, 1]. Sum of this array of prefix minimum is 5 (= Y).

方法：解决此问题的方法基于以下思想：

Suppose the remaining array to be created at some point of time be len and the remaining sum to be rem.

Greedily choose a value for this index by the given method below:

• let’s take Z value at this index, then we should have at least (Z + len – 1) = rem (By taking 1 at remaining indices).
• So, we get Z = (rem + 1 – len). Now, this value might be greater than len but we can’t take it. So, we will choose min(Z, len).

请按照以下步骤解决此问题：

• 现在，已经从上述贪心方法构建了所需排列的前缀最小数组。
• 它也可能有重复。因此，要以相反的顺序删除该数组的迭代，并且每当 arr[i] = arr[i-1] 时，将不存在于任何索引处的最小元素放在第 i^个索引处。
• 这样，确保前缀最小值之和应为 Y，并且创建的数组也应为排列。
• 打印最终数组

下面是上述方法的实现：

C++

// C++ program for Generate permutation
// of length N such that sum of all prefix
// minimum of that permutation is Y.
#include 
#include 
#include 
using namespace std;
 
// Find the value greedily for the
// current index  as discussed in approach
int findValue(long long N, long long Y)
{
    return min(N, Y + 1 - N);
}
 
// Function to generate the permutation
void generatePermutation(long long N, long long Y)
{
    // Store the prefix minimum array first
    // then we will convert it to permutation
    vector ans(N);
 
    // If Y should belong to [N, (N*(N + 1)/2)],
    // otherwise we will print -1;
    if (Y < N || (2 * Y) > (N * (N + 1))) {
        cout << -1 << endl;
        return;
    }
 
    // Remaining elements to be taken
    set s;
 
    for (int i = 1; i <= N; i++) {
        s.insert(i);
    }
 
    // Generate prefix minimum array
    for (int i = 0; i < N; i++) {
        // Length remaining
        int len = N - i;
        int val = findValue(len, Y);
        ans[i] = val;
        Y -= val;
        if (s.find(val) != s.end())
            s.erase(val);
    }
 
    // Remove duplicates to make array permutation
    // So, iterate in reverse order
    for (int i = N - 1; i > 0; i--) {
        if (ans[i] == ans[i - 1]) {
            // Find minimum element not taken
            // in the permutation
            ans[i] = *s.begin();
            s.erase(ans[i]);
        }
    }
 
    // Print the permutation
    for (auto i : ans) {
        cout << i << " ";
    }
    cout << endl;
}
 
// Driver Code
int main()
{
 
    long long N = 5, Y = 10;
 
    generatePermutation(N, Y);
    return 0;
}

Java

// Java program for Generate permutation
// of length N such that sum of all prefix
// minimum of that permutation is Y.
import java.util.*;
 
class GFG {
 
  // Find the value greedily for the
  // current index  as discussed in approach
  static int findValue(int  N, int  Y)
  {
    return Math.min(N, Y + 1 - N);
  }
 
  // Function to generate the permutation
  static void generatePermutation( int N,  int Y)
  {
    // Store the prefix minimum array first
    // then we will convert it to permutation
    int[] ans = new int[N];
 
    // If Y should belong to [N, (N*(N + 1)/2)],
    // otherwise we will print -1;
    if (Y < N || (2 * Y) > (N * (N + 1))) {
      System.out.println(-1);
      return;
    }
 
    // Remaining elements to be taken
    Set s = new HashSet();
 
    for (int i = 1; i <= N; i++) {
      s.add(i);
    }
 
    // Generate prefix minimum array
    for (int i = 0; i < N; i++) {
      // Length remaining
      int len = N - i;
      int val = findValue(len, Y);
      ans[i] = val;
      Y -= val;
      if (s.contains(val))
        s.remove(val);
    }
 
    // Remove duplicates to make array permutation
    // So, iterate in reverse order
    for (int i = N - 1; i > 0; i--) {
      if (ans[i] == ans[i - 1]) {
        // Find minimum element not taken
        // in the permutation
        ans[i] = s.stream().findFirst().get();
        s.remove(ans[i]);
      }
    }
 
    // Print the permutation
    for (int i = 0; i < N; i++) {
      System.out.print(ans[i] + " ");
    }
  }
 
  // Driver Code
  public static void main (String[] args) {
 
    int N = 5, Y = 10;
 
    generatePermutation(N, Y);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# python3 program for Generate permutation
# of length N such that sum of all prefix
# minimum of that permutation is Y.
 
# Find the value greedily for the
# current index as discussed in approach
def findValue(N, Y):
    return min(N, Y + 1 - N)
 
# Function to generate the permutation
def generatePermutation(N, Y):
 
    # Store the prefix minimum array first
    # then we will convert it to permutation
    ans = [0 for _ in range(N)]
 
    # If Y should belong to [N, (N*(N + 1)/2)],
    # otherwise we will print -1;
    if (Y < N or (2 * Y) > (N * (N + 1))):
        print(-1)
        return
 
    # Remaining elements to be taken
    s = set()
 
    for i in range(1, N+1):
        s.add(i)
 
    # Generate prefix minimum array
    for i in range(0, N):
        # Length remaining
        len = N - i
        val = findValue(len, Y)
        ans[i] = val
        Y -= val
        if (val in s):
            s.remove(val)
 
    # Remove duplicates to make array permutation
    # So, iterate in reverse order
    for i in range(N-1, -1, -1):
        if (ans[i] == ans[i - 1]):
            # Find minimum element not taken
            # in the permutation
            ans[i] = list(s)[0]
            s.remove(ans[i])
 
    # Print the permutation
    for i in ans:
        print(i, end=" ")
 
    print()
 
# Driver Code
if __name__ == "__main__":
 
    N, Y = 5, 10
 
    generatePermutation(N, Y)
 
# This code is contributed by rakeshsahni

C#

// C# implementation of above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG{
 
  // Find the value greedily for the
  // current index  as discussed in approach
  static int findValue(int  N, int  Y)
  {
    return Math.Min(N, Y + 1 - N);
  }
 
  // Function to generate the permutation
  static void generatePermutation( int N,  int Y)
  {
    // Store the prefix minimum array first
    // then we will convert it to permutation
    int[] ans = new int[N];
 
    // If Y should belong to [N, (N*(N + 1)/2)],
    // otherwise we will print -1;
    if (Y < N || (2 * Y) > (N * (N + 1))) {
      Console.Write(-1);
      return;
    }
 
    // Remaining elements to be taken
    HashSet s = new HashSet();
 
    for (int i = 1; i <= N; i++) {
      s.Add(i);
    }
 
    // Generate prefix minimum array
    for (int i = 0; i < N; i++) {
      // Length remaining
      int len = N - i;
      int val = findValue(len, Y);
      ans[i] = val;
      Y -= val;
      if (s.Contains(val))
        s.Remove(val);
    }
 
    // Remove duplicates to make array permutation
    // So, iterate in reverse order
    for (int i = N - 1; i > 0; i--) {
      if (ans[i] == ans[i - 1]) {
        // Find minimum element not taken
        // in the permutation
        ans[i] = s.First();
        s.Remove(ans[i]);
      }
    }
 
    // Print the permutation
    for (int i = 0; i < N; i++) {
      Console.Write(ans[i] + " ");
    }
  }
 
  // Driver Code
  static public void Main (){
 
    int N = 5, Y = 10;
 
    generatePermutation(N, Y);
  }
}
 
// This code is contributed by code_hunt.

Javascript

 
// JavaScript program for Generate permutation
// of length N such that sum of all prefix
// minimum of that permutation is Y.
 
// Find the value greedily for the
// current index  as discussed in approach
function findValue(N, Y)
{
    return Math.min(N, Y + 1 - N);
}
 
// Function to generate the permutation
function generatePermutation(N, Y)
{
    // Store the prefix minimum array first
    // then we will convert it to permutation
    let ans = new Array(N);
 
    // If Y should belong to [N, (N*(N + 1)/2)],
    // otherwise we will print -1;
    if (Y < N || (2 * Y) > (N * (N + 1))) {
        document.write(-1,"</br>");
        return;
    }
 
    // Remaining elements to be taken
    let s = new Set();
 
    for (let i = 1; i <= N; i++) {
        s.add(i);
    }
 
    // Generate prefix minimum array
    for (let i = 0; i <N; i++) {
        // Length remaining
        let len = N - i;
        let val = findValue(len, Y);
        ans[i] = val;
        Y -= val;
        if (s.has(val))
            s.delete(val);
    }
 
    // Remove duplicates to make array permutation
    // So, iterate in reverse order
    for (let i = N - 1; i > 0; i--) {
        if (ans[i] == ans[i - 1]) {
            // Find minimum element not taken
            // in the permutation
            ans[i] = Array.from(s)[0]
            s.delete(ans[i]);
        }
    }
 
    // Print the permutation
    for (let i of ans) {
        document.write(i," ");
    }
    document.write("</N;>");
}
 
// Driver Code
let N = 5, Y = 10;
generatePermutation(N, Y);
 
// This code is contributed by shinjanpatra
 

输出

5 2 1 4 3 

时间复杂度： O(N * log N)。
如果我们使用一个集合，我们可以通过使用两个指针技术使其成为 O(N)。
辅助空间： O(N)