查找给定数组中的所有子数组索引范围，设置位和等于 X

给定一个长度为N的数组arr （从 1 开始的索引）和一个整数X ，任务是查找并打印数组中所有设置位和等于X的索引范围。

例子：

Input: A[] = {1 4 3 5 7}, X = 4
Output:  (1, 3), (3, 4)
Explanation: In the above array subarray having set bit sum equal to X (= 4). 
Starting from index 1 to 3. {1 4 3}  = (001) + (100) + (011) = 4  and 
other one is from 3 to 4 {3, 5} = (011) + (101) = 4.

Input: arr[] = {5, 3, 0,  4, 10}, X = 7
Output:  (1 5)
Explanation: In the above array  subarrays having set bit sum equal to X(= 7) start from 1 to 5 only.

方法：使用两个指针方法解决问题。

• 编写一个函数countSetBit 来计算设置的位数。
  • 初始化计数器 c=0，以存储数组中每个数字的单独计数。
  • 遍历数组并检查每个设置位并增加计数器。
  • 用一组设置位的计数替换每个数字
• 编写一个函数来打印一系列子数组 PrintIndex
  使用两个指针 i 和 j 运行循环并检查总和，如下所示：
  • 如果当前索引总和小于 X，则将 arr[j] 处的值添加到 currsum
  • 否则，如果总和等于 X，则推回数组的开始和结束索引并增加计数器 i。
  • 否则递减计数器，从 currsum 中减去 arr[i] 处的值。
  • 对所有元素重复相同的操作。
    

下面是上述方法的实现：

C++

// C++ program to Find all range
// Having set bit sum X  in array
#include 
using namespace std;
 
// Function to replace elements
// With their set bit count
void countSetBit(vector& arr, int n)
{
    int c = 0, i;
 
    for (i = 0; i < n; i++) {
 
        int x = arr[i];
        while (x) {
            int l = x % 10;
            if (x & 1)
                c++;
            x /= 2;
        }
        // Replace array element
        // to set bit count
        arr[i] = c;
        c = 0;
    }
}
 
// Function to find range of subarrays
// having set bit sum equal to X.
void PrintIndex(vector arr, int N, int X,
                vector& v)
{
 
    int i = 0, j = 0, currSum = arr[0];
    while (j < N && i < N) {
        if (currSum == X) {
            // push back index i start
            // point ans end point j
            // when sum == X
 
            v.push_back(i + 1);
            v.push_back(j + 1);
 
            j++;
            currSum += arr[j];
        }
        // when current sum is
        // less than X increment j
        // and add arr[j]
        else if (currSum < X) {
            j++;
            currSum += arr[j];
        }
        // when current sum is
        // greater than X increment j
        // and subtract arr[i]
        else {
            currSum -= arr[i];
            i++;
        }
    }
}
 
// Driver code
int main()
{
    vector v = { 1, 4, 3, 5, 7 };
    int X = 4;
    int N = v.size();
 
    // replace all the array element into
    // their set bit count value
    countSetBit(v, N);
 
    vector ans;
 
    PrintIndex(v, N, X, ans);
 
    for (int i = 0; i < ans.size() - 1; i += 2)
        cout << "(" << ans[i] << " "
             << ans[i + 1] << ")"
             << " ";
 
    return 0;
}

Java

// JAVA code to implement the above approach
import java.util.*;
class GFG {
 
  // Function to replace elements
  // With their set bit count
  static void countSetBit(int[] arr, int n)
  {
    int c = 0, i;
 
    for (i = 0; i < n; i++) {
 
      int x = arr[i];
      while (x > 0) {
        int l = x % 10;
        if ((x & 1) == 1)
          c++;
        x /= 2;
      }
      // Replace array element
      // to set bit count
      arr[i] = c;
      c = 0;
    }
  }
 
  // Function to find range of subarrays
  // having set bit sum equal to X.
  static void PrintIndex(int[] arr, int N, int X,
                         ArrayList v)
  {
 
    int i = 0, j = 0, currSum = arr[0];
    while (j < N && i < N) {
      if (currSum == X) {
        // push back index i start
        // point ans end point j
        // when sum == X
 
        v.add(i + 1);
        v.add(j + 1);
 
        j++;
        if (j < N)
          currSum += arr[j];
      }
      // when current sum is
      // less than X increment j
      // and add arr[j]
      else if (currSum < X) {
        j++;
        if (j < N)
          currSum += arr[j];
      }
      // when current sum is
      // greater than X increment j
      // and subtract arr[i]
      else {
        currSum -= arr[i];
        i++;
      }
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int[] v = { 1, 4, 3, 5, 7 };
    int X = 4;
    int N = v.length;
 
    // replace all the array element into
    // their set bit count value
    countSetBit(v, N);
 
    ArrayList ans = new  ArrayList();
 
    PrintIndex(v, N, X, ans);
 
    for (int i = 0; i < ans.size() - 1; i += 2)
      System.out.print("(" + ans.get(i) + " " + ans.get(i + 1)
                       + ")"
                       + " ");
  }
}
 
// This code is contributed by sanjoy_62.

C#

// C# program to Find all range
// Having set bit sum X  in array
using System;
using System.Collections;
 
class GFG {
 
  // Function to replace elements
  // With their set bit count
  static void countSetBit(int[] arr, int n)
  {
    int c = 0, i;
 
    for (i = 0; i < n; i++) {
 
      int x = arr[i];
      while (x > 0) {
        int l = x % 10;
        if ((x & 1) == 1)
          c++;
        x /= 2;
      }
      // Replace array element
      // to set bit count
      arr[i] = c;
      c = 0;
    }
  }
 
  // Function to find range of subarrays
  // having set bit sum equal to X.
  static void PrintIndex(int[] arr, int N, int X,
                         ArrayList v)
  {
 
    int i = 0, j = 0, currSum = arr[0];
    while (j < N && i < N) {
      if (currSum == X) {
        // push back index i start
        // point ans end point j
        // when sum == X
 
        v.Add(i + 1);
        v.Add(j + 1);
 
        j++;
        if (j < N)
          currSum += arr[j];
      }
      // when current sum is
      // less than X increment j
      // and add arr[j]
      else if (currSum < X) {
        j++;
        if (j < N)
          currSum += arr[j];
      }
      // when current sum is
      // greater than X increment j
      // and subtract arr[i]
      else {
        currSum -= arr[i];
        i++;
      }
    }
  }
 
  // Driver code
  public static void Main()
  {
    int[] v = { 1, 4, 3, 5, 7 };
    int X = 4;
    int N = v.Length;
 
    // replace all the array element into
    // their set bit count value
    countSetBit(v, N);
 
    ArrayList ans = new ArrayList();
 
    PrintIndex(v, N, X, ans);
 
    for (int i = 0; i < ans.Count - 1; i += 2)
      Console.Write("(" + ans[i] + " " + ans[i + 1]
                    + ")"
                    + " ");
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program to Find all range
    // Having set bit sum X in array
 
    // Function to replace elements
    // With their set bit count
    const countSetBit = (arr, n) => {
        let c = 0, i;
 
        for (i = 0; i < n; i++) {
 
            let x = arr[i];
            while (x) {
                let l = x % 10;
                if (x & 1)
                    c++;
                x = parseInt(x / 2);
            }
             
            // Replace array element
            // to set bit count
            arr[i] = c;
            c = 0;
        }
    }
 
    // Function to find range of subarrays
    // having set bit sum equal to X.
    const PrintIndex = (arr, N, X, v) => {
 
        let i = 0, j = 0, currSum = arr[0];
        while (j < N && i < N)
        {
            if (currSum == X)
            {
             
                // push back index i start
                // point ans end point j
                // when sum == X
 
                v.push(i + 1);
                v.push(j + 1);
 
                j++;
                currSum += arr[j];
            }
             
            // when current sum is
            // less than X increment j
            // and add arr[j]
            else if (currSum < X) {
                j++;
                currSum += arr[j];
            }
             
            // when current sum is
            // greater than X increment j
            // and subtract arr[i]
            else {
                currSum -= arr[i];
                i++;
            }
        }
    }
 
    // Driver code
    let v = [1, 4, 3, 5, 7];
    let X = 4;
    let N = v.length;
 
    // replace all the array element into
    // their set bit count value
    countSetBit(v, N);
 
    let ans = [];
 
    PrintIndex(v, N, X, ans);
 
    for (let i = 0; i < ans.length - 1; i += 2)
        document.write(`(${ans[i]} ${ans[i + 1]}) `);
 
// This code is contributed by rakeshsahni
 

输出：

(1 3) (3 4)

时间复杂度： O(N * d) 其中 d 是数组元素中的位数
辅助空间： O(N)