查找具有最大值和最小值之间差异的子数组范围正好 K

给定一个长度为N和整数K的数组arr[] ，任务是打印数组的子数组范围（起始索引，结束索引），其中子数组的最大和最小元素之间的差恰好是 K。（基于 1 的索引）

例子：

Input: arr[] = {2, 1, 3, 4, 2, 6}, K = 2
Output: (1, 3), (2, 3), (3, 5), (4, 5)
Explanation: In the above array following sub array ranges have max min difference exactly K
(1, 3) => max = 3 and min = 1. Difference = 3 – 1 = 2
(2, 3) => max = 3 and min = 1. Difference = 3 – 1 = 2
(3, 5) => max = 4 and min = 2. Difference = 4 – 2 = 2
(4, 5) => max = 4 and min = 2. Difference = 4 – 2 = 2

Input: arr[] = {5, 3, 4, 6, 1, 2}, K = 6
Output: -1
Explanation: There is no such sub array ranges.

编程需要懂一点英语

方法：解决问题的基本思想是形成所有子数组，并找到每个子数组的最小值和最大值及其差异。请按照以下步骤解决问题。

从i = 0 到 N-1遍历数组：
- 从j = i 迭代到 N-1 ：
  - 在存储从i开始的当前子数组的集合中插入arr[j]当前元素。
  - 找到集合的最小值和最大值。
  - 获取它们的差值并检查它们的差值是否等于K。
  - 如果是，则在答案中推动这个范围。
返回范围。
如果没有这样的范围，则返回 -1。

下面是上述方法的实现。

C++

// C++ implementation of above approach
 
#include 
using namespace std;
 
// Function to print index ranges
void printRanges(vector arr, int n,
                 int K)
{
    int i, j, f = 0;
    for (i = 0; i < n; i++) {
 
        // Set to store the elements
        // of a subarray
        set s;
        for (j = i; j < n; j++) {
 
            // Insert current element in set
            s.insert(arr[j]);
 
            // Calculate max and min for
            // any particular index range
            int max = *s.rbegin();
            int min = *s.begin();
 
            // If we get max-min = K
            // print 1 based index
            if (max - min == K) {
                cout << i + 1 << " " << j + 1
                     << "\n";
                f = 1;
            }
        }
    }
 
    // If we didn't find any index ranges
    if (f == 0)
        cout << -1 << endl;
}
 
// Driver Code
int main()
{
    vector arr = { 2, 1, 3, 4, 2, 6 };
    int N = arr.size();
    int K = 2;
 
    // Function call
    printRanges(arr, N, K);
    return 0;
}

Java

// Java implementation of above approach
import java.util.*; 
 
class GFG {
 
  // Function to print index ranges
  static void printRanges(int arr[], int n,
                          int K)
  {
    int i, j, f = 0;
    for (i = 0; i < n; i++) {
 
      // Set to store the elements
      // of a subarray
      Set s = new HashSet<>();
      for (j = i; j < n; j++) {
 
        // Insert current element in set
        s.add(arr[j]);
 
        // Calculate max and min for
        // any particular index range
        int max = Collections.max(s);
        int min = Collections.min(s);
 
        // If we get max-min = K
        // print 1 based index
        if (max - min == K) {
          System.out.println( (i + 1) + " " + (j + 1));
          f = 1;
        }
      }
    }
 
    // If we didn't find any index ranges
    if (f == 0)
      System.out.println(-1);
  }
 
  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 2, 1, 3, 4, 2, 6 };
    int N = arr.length;
    int K = 2;
 
    // Function call
    printRanges(arr, N, K);
  }
}
 
// This code is contributed by hrithikgarg03188.

输出

时间复杂度： O(N ² * logN)
辅助空间： O(N)