Given two integer array, L and R and an integer N. Each range from the given array denote that every number in the range [L[i], R[i]] is present. The task is to calculate the N-th(0-based indexing) element when numbers from given ranges are ordered in their sorted order.
Examples:
Input: L = {1, 5}, R = {3, 7}, N = 4
Output: 6
Explanation: The numbers present are {1, 2, 3, 5, 6, 7}. Therefore 4-th element(0-indexed) is 6.Input: L = {1, 3}, R = {4, 5}, N = 3
Output: 3
Explanation: The numbers present are {1, 2, 3, 4, 3, 4, 5} and their sorted order is {1, 2, 3, 3, 4, 4, 5}. Therefore 3-rd element(0-indexed) is 3.
Approach: The task can be solved with the help of a binary search over the answer.
Follow the below steps to solve the problem:
- Calculate two variables min and max which store the minimum element from the L array and maximum element from the R array.
- The range of the binary search will be [min, max].
- For each mid = (min + max) / 2 calculate the position of the current element.
- To calculate the position, iterate over all ranges and make a variable t = 0 to store the position. Check below two conditions if L[i] >= mid
- If mid <= R[i], update t += mid – L[i] + 1.
- else t += R[i] – L[i] + 1
- The binary search range can be updated as
- if(t > n) max = mid – 1.
- else min = mid + 1.
- The final answer will be stored in the variable min.
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;int nthElement(vector<int> L, vector<int> R, int n){ // Store the size of the ranges int K = L.size(); // Calculate the max and min values long long min = 2000000000, max = -2000000000; for (int i = 0; i < K; i++) { if (L[i] < min) min = L[i]; if (R[i] > max) max = R[i]; } // Do a binary search over answer while (min <= max) { long long mid = (min + max) / 2; long long t = 0; for (int i = 0; i < K; i++) { if (mid >= L[i]) { if (mid <= R[i]) { t += mid - L[i] + 1; } else { t += R[i] - L[i] + 1; } } } // Update the binary Search range. if (t > n) { max = mid - 1; } else { min = mid + 1; } } return min;}// Driver Codeint main(){ vector<int> L = { 1, 5 }, R = { 3, 7 }; int N = 4; cout << nthElement(L, R, N);} |
Java
// Java implementation for the above approachclass GFG { public static long nthElement(int[] L, int[] R, int n) { // Store the size of the ranges int K = L.length; // Calculate the max and min values long min = 2000000000, max = -2000000000; for (int i = 0; i < K; i++) { if (L[i] < min) min = L[i]; if (R[i] > max) max = R[i]; } // Do a binary search over answer while (min <= max) { long mid = (min + max) / 2; long t = 0; for (int i = 0; i < K; i++) { if (mid >= L[i]) { if (mid <= R[i]) { t += mid - L[i] + 1; } else { t += R[i] - L[i] + 1; } } } // Update the binary Search range. if (t > n) { max = mid - 1; } else { min = mid + 1; } } return min; } // Driver Code public static void main(String args[]) { int[] L = { 1, 5 }, R = { 3, 7 }; int N = 4; System.out.println(nthElement(L, R, N)); }}// This code is contributed by gfgking |
Python3
# Python implementation for the above approachdef nthElement(L, R, n): # Store the size of the ranges K = len(L) # Calculate the max and min values min = 2000000000 max = -2000000000; for i in range(K): if (L[i] < min): min = L[i] if (R[i] > max): max = R[i]; # Do a binary search over answer while (min <= max): mid = (min + max) // 2; t = 0; for i in range(K): if (mid >= L[i]): if (mid <= R[i]): t += mid - L[i] + 1; else: t += R[i] - L[i] + 1; # Update the binary Search range. if (t > n): max = mid - 1; else: min = mid + 1; return min;# Driver CodeL = [1, 5]R = [3, 7];N = 4;print(nthElement(L, R, N));# This code is contributed by gfgking |
C#
// C# implementation for the above approachusing System;class GFG { public static long nthElement(int[] L, int[] R, int n) { // Store the size of the ranges int K = L.Length; // Calculate the max and min values long min = 2000000000, max = -2000000000; for (int i = 0; i < K; i++) { if (L[i] < min) min = L[i]; if (R[i] > max) max = R[i]; } // Do a binary search over answer while (min <= max) { long mid = (min + max) / 2; long t = 0; for (int i = 0; i < K; i++) { if (mid >= L[i]) { if (mid <= R[i]) { t += mid - L[i] + 1; } else { t += R[i] - L[i] + 1; } } } // Update the binary Search range. if (t > n) { max = mid - 1; } else { min = mid + 1; } } return min; } // Driver Code public static void Main() { int[] L = { 1, 5 }, R = { 3, 7 }; int N = 4; Console.Write(nthElement(L, R, N)); }}// This code is contributed by gfgking |
Javascript
<script>// Javascript implementation for the above approachfunction nthElement(L, R, n) { // Store the size of the ranges let K = L.length; // Calculate the max and min values let min = 2000000000, max = -2000000000; for (let i = 0; i < K; i++) { if (L[i] < min) min = L[i]; if (R[i] > max) max = R[i]; } // Do a binary search over answer while (min <= max) { let mid = Math.floor((min + max) / 2); let t = 0; for (let i = 0; i < K; i++) { if (mid >= L[i]) { if (mid <= R[i]) { t += mid - L[i] + 1; } else { t += R[i] - L[i] + 1; } } } // Update the binary Search range. if (t > n) { max = mid - 1; } else { min = mid + 1; } } return min;}// Driver Codelet L = [1, 5]let R = [3, 7];let N = 4;document.write(nthElement(L, R, N));// This code is contributed by gfgking</script> |
Output
6
Time Complexity: O(N*log(max – min))
Auxiliary Space: O(N)
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