包含 Q 次运算后所有唯一元素的最小长度子数组

给定一个大小为N的数组，最初包含所有元素为0 ，以及包含[L, R]形式的范围的Q查询。任务是通过向Q查询范围[L, R]中的每个元素添加1来修改数组，然后打印包含所有唯一元素的最小长度子数组的大小。
注意：在 [L, R] 范围内使用基于 1 的索引。
例子：

Input: N = 6, Q[4][] = {{1, 3}, {4, 6}, {3, 4}, {3, 3}} 
Output: 3 
Explanation: 
Initial array: arr[] = { 0, 0, 0, 0, 0, 0 } 
Query 1: updated arr[] = { 1, 1, 1, 0, 0, 0 }. 
Query 2: updated arr[] = { 1, 1, 1, 1, 1, 1 }. 
Query 3: updated arr[] = { 1, 1, 2, 2, 1, 1 }. 
Query 4: updated arr[] = { 1, 1, 3, 2, 1, 1 }. 
The subarray { 1, 3, 2 } is minimum subarray which contains all unique elements. Thus, the answer is 3.
Input: N = 8, Q[6][] = {{1, 4}, {3, 4}, {4, 5}, {5, 5}, {7, 8}, {8, 8}} 
Output: 4 
Explanation: 
After processing all queries, the array becomes = { 1, 1, 2, 3, 2, 0, 1, 2 }. 
The subarray { 3, 2, 0, 1 } is minimum subarray which contains all unique elements. Thus, the answer is 4.

方法：这个想法是使用Prefix sum数组的概念和两个指针的方法来解决这个问题。

• 查询后的最终数组可以通过在索引L处将数组中的值增加 1 并在索引R + 1处将值减少1 来计算。

Processing of a Query:
arr[L] += 1
arr[R + 1] -= 1

• 最后，计算数组的前缀和以计算查询后的最终数组。然后使用两个指针的方法得到包含所有唯一元素的最小长度子数组。

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges
 
#include 
using namespace std;
 
// Function to find minimum size subarray
// of all array elements
int subarrayLength(int A[], int R[][2],
                   int N, int M)
{
 
    // Updating the array after
    // processing each query
    for (int i = 0; i < M; ++i) {
 
        int l = R[i][0], r = R[i][1] + 1;
 
        // Making it to 0-indexing
        l--;
        r--;
 
        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;
 
        if (r < N)
            A[r]--;
    }
 
    // Iterating over the array
    // to get the final array
    for (int i = 1; i < N; ++i) {
        A[i] += A[i - 1];
    }
 
    // Variable to get count
    // of all unique elements
    int count = 0;
 
    // Hash to maintain perviously
    // occurred elements
    unordered_set s;
 
    // Loop to find the all
    // unique elements
    for (int i = 0; i < N; ++i) {
        if (s.find(A[i]) == s.end())
            count++;
 
        s.insert(A[i]);
    }
 
    // array to maintain counter
    // of encountered elements
    vector repeat(count + 1, 0);
 
    // variable to store answer
    int ans = N;
 
    // Using two pointers approach
    int counter = 0, left = 0, right = 0;
 
    while (right < N) {
 
        int cur_element = A[right];
        repeat[cur_element] += 1;
 
        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;
 
        // when all unique
        // elements are found
        while (counter == count) {
 
            // update answer with
            // minimum size
            ans = min(ans, right - left + 1);
 
            // decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;
 
            // decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }
 
        ++right;
    }
    return ans;
}
 
// Driver code
int main()
{
    int N = 8, queries = 6;
    int Q[][2]
        = {
            { 1, 4 }, { 3, 4 }, { 4, 5 },
            { 5, 5 }, { 7, 8 }, { 8, 8 }
          };
 
    int A[N] = { 0 };
 
    cout << subarrayLength(A, Q, N, queries);
 
    return 0;
}

Java

// Java implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges
import java.util.*;
class GFG{
 
// Function to find minimum size subarray
// of all array elements
static int subarrayLength(int A[], int R[][],
                          int N, int M)
{
    // Updating the array after
    // processing each query
    for (int i = 0; i < M; ++i)
    {
        int l = R[i][0], r = R[i][1] + 1;
 
        // Making it to 0-indexing
        l--;
        r--;
 
        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;
 
        if (r < N)
            A[r]--;
    }
 
    // Iterating over the array
    // to get the final array
    for (int i = 1; i < N; ++i)
    {
        A[i] += A[i - 1];
    }
 
    // Variable to get count
    // of all unique elements
    int count = 0;
 
    // Hash to maintain perviously
    // occurred elements
    HashSet s = new HashSet();
 
    // Loop to find the all
    // unique elements
    for (int i = 0; i < N; ++i)
    {
        if (!s.contains(A[i]))
            count++;
        s.add(A[i]);
    }
 
    // array to maintain counter
    // of encountered elements
    int []repeat = new int[count + 1];
 
    // variable to store answer
    int ans = N;
 
    // Using two pointers approach
    int counter = 0, left = 0, right = 0;
 
    while (right < N)
    {
        int cur_element = A[right];
        repeat[cur_element] += 1;
 
        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;
 
        // when all unique
        // elements are found
        while (counter == count)
        {
            // update answer with
            // minimum size
            ans = Math.min(ans,
                           right - left + 1);
 
            // decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;
 
            // decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }
 
        ++right;
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 8, queries = 6;
    int Q[][] = {{ 1, 4 }, { 3, 4 }, { 4, 5 },
                 { 5, 5 }, { 7, 8 }, { 8, 8 }};
    int A[] = new int[N];
    System.out.print(subarrayLength(A, Q,
                                    N, queries));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 implementation to find the
# minimum size subarray containing
# all unique elements after processing
# the array for K queries of ranges
 
# Function to find minimum size subarray
# of all array elements
def subarrayLength(A, R, N, M):
 
    # Updating the array after
    # processing each query
    for i in range(M):
 
        l = R[i][0]
        r = R[i][1] + 1
 
        # Making it to 0-indexing
        l -= 1
        r -= 1
 
        # Prefix sum array concept is used
        # to obtain the count array
        A[l] += 1
 
        if (r < N):
            A[r] -= 1
 
    # Iterating over the array
    # to get the final array
    for i in range(1 , N):
        A[i] += A[i - 1]
 
    # Variable to get count
    # of all unique elements
    count = 0
 
    # Hash to maintain perviously
    # occurred elements
    s = []
 
    # Loop to find the all
    # unique elements
    for i in range(N):
        if (A[i] not in s):
            count += 1
 
        s.append(A[i])
 
    # Array to maintain counter
    # of encountered elements
    repeat = [0] * (count + 1)
 
    # Variable to store answer
    ans = N
 
    # Using two pointers approach
    counter, left, right = 0, 0, 0
 
    while (right < N):
 
        cur_element = A[right]
        repeat[cur_element] += 1
 
        # Increment counter
        # if occurred once
        if (repeat[cur_element] == 1):
            counter += 1
 
        # When all unique
        # elements are found
        while (counter == count):
 
            # update answer with
            # minimum size
            ans = min(ans, right - left + 1)
 
            # Decrement count of
            # elements from left
            cur_element = A[left]
            repeat[cur_element] -= 1
            left += 1
 
            # Decrement counter
            if (repeat[cur_element] == 0):
                counter -= 1
                 
        right += 1
         
    return ans
 
# Driver code
if __name__ == "__main__":
     
    N , queries = 8 , 6
    Q = [ [ 1, 4 ], [ 3, 4 ], [ 4, 5 ],
          [ 5, 5 ], [ 7, 8 ], [ 8, 8 ] ]
 
    A = [0] * N
    print(subarrayLength(A, Q, N, queries))
 
# This code is contributed by chitranayal

C#

// C# implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find minimum size subarray
// of all array elements
static int subarrayLength(int []A, int [,]R,
                          int N, int M)
{
     
    // Updating the array after
    // processing each query
    for(int i = 0; i < M; ++i)
    {
        int l = R[i, 0], r = R[i, 1] + 1;
 
        // Making it to 0-indexing
        l--;
        r--;
 
        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;
 
        if (r < N)
            A[r]--;
    }
 
    // Iterating over the array
    // to get the readonly array
    for(int i = 1; i < N; ++i)
    {
        A[i] += A[i - 1];
    }
 
    // Variable to get count
    // of all unique elements
    int count = 0;
 
    // Hash to maintain perviously
    // occurred elements
    HashSet s = new HashSet();
 
    // Loop to find the all
    // unique elements
    for(int i = 0; i < N; ++i)
    {
        if (!s.Contains(A[i]))
            count++;
             
        s.Add(A[i]);
    }
 
    // Array to maintain counter
    // of encountered elements
    int []repeat = new int[count + 1];
 
    // Variable to store answer
    int ans = N;
 
    // Using two pointers approach
    int counter = 0, left = 0, right = 0;
 
    while (right < N)
    {
        int cur_element = A[right];
        repeat[cur_element] += 1;
 
        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;
 
        // When all unique
        // elements are found
        while (counter == count)
        {
             
            // Update answer with
            // minimum size
            ans = Math.Min(ans,
                           right - left + 1);
 
            // Decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;
 
            // Decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }
        ++right;
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int N = 8, queries = 6;
    int [,]Q = { { 1, 4 }, { 3, 4 }, { 4, 5 },
                 { 5, 5 }, { 7, 8 }, { 8, 8 } };
    int []A = new int[N];
     
    Console.Write(subarrayLength(A, Q,
                                 N, queries));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

 
// Javascript implementation to find the
// minimum size subarray containing
// all unique elements after processing
// the array for K queries of ranges
 
// Function to find minimum size subarray
// of all array elements
function subarrayLength(A, R, N, M)
{
    // Updating the array after
    // processing each query
    for (let i = 0; i < M; ++i)
    {
        let l = R[i][0], r = R[i][1] + 1;
   
        // Making it to 0-indexing
        l--;
        r--;
   
        // Prefix sum array concept is used
        // to obtain the count array
        A[l]++;
   
        if (r < N)
            A[r]--;
    }
   
    // Iterating over the array
    // to get the final array
    for (let i = 1; i < N; ++i)
    {
        A[i] += A[i - 1];
    }
   
    // Variable to get count
    // of all unique elements
    let count = 0;
   
    // Hash to maintain perviously
    // occurred elements
    let s = new Set();
   
    // Loop to find the all
    // unique elements
    for (let i = 0; i < N; ++i)
    {
        if (!s.has(A[i]))
            count++;
        s.add(A[i]);
    }
   
    // array to maintain counter
    // of encountered elements
    let repeat = Array.from({length: count + 1}, (_, i) => 0);
   
    // variable to store answer
    let ans = N;
   
    // Using two poleters approach
    let counter = 0, left = 0, right = 0;
   
    while (right < N)
    {
        let cur_element = A[right];
        repeat[cur_element] += 1;
   
        // Increment counter
        // if occurred once
        if (repeat[cur_element] == 1)
            ++counter;
   
        // when all unique
        // elements are found
        while (counter == count)
        {
            // update answer with
            // minimum size
            ans = Math.min(ans,
                           right - left + 1);
   
            // decrement count of
            // elements from left
            cur_element = A[left];
            repeat[cur_element] -= 1;
            ++left;
   
            // decrement counter
            if (repeat[cur_element] == 0)
                --counter;
        }
   
        ++right;
    }
    return ans;
}
 
// Driver code
     
      let N = 8, queries = 6;
    let Q = [[ 1, 4 ], [ 3, 4 ], [ 4, 5 ],
                 [ 5, 5 ], [ 7, 8 ], [ 8, 8 ]];
    let A = Array.from({length: N}, (_, i) => 0);
    document.write(subarrayLength(A, Q,
                                    N, queries));
                                                                                      

输出：

4

时间复杂度： O(N)

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