Generating subarrays using recursion
Given an array, generate all the possible subarrays of the given array using recursion.
Examples:
Input : [1, 2, 3] Output : [1], [1, 2],[1,3], [2], [1, 2, 3], [2, 3], [3] Input : [1, 2] Output : [1], [1, 2], [2]
Approach: We use two pointers start and end to maintain the starting and ending point of the array and follow the steps given below:
- Stop if we have reached the end of the array
- Increment the end index if start has become greater than end
- Print the subarray from index start to end and increment the starting index
Below is the implementation of the above approach.
// C++ code to print all possible subarrays // for given array using recursion#include <iostream># include <vector>using namespace std;// Recursive function to print all possible subarrays // for given arrayvoid printSubArrays(vector<int> arr, int start, int end){ // Stop if we have reached the end of the array if (end == arr.size()) return; // Increment the end point and start from 0 else if (start > end) printSubArrays(arr, 0, end + 1); // Print the subarray and increment the starting point else { cout << "["; for (int i = start; i < end; i++){ cout << arr[i] << ", "; } cout << arr[end] << "]" << endl; printSubArrays(arr, start + 1, end); } return;}int main(){ vector<int> arr = {1, 2, 3}; printSubArrays(arr, 0, 0); return 0;} |
Output:
[1] [1, 2] [2] [1, 2, 3] [2, 3] [3]
Time Complexity: 
Last Updated on November 13, 2021 by admin