Given an array arr[] consisting of non-negative integers and an integer k. The task is to find the minimum length of any sub-array of arr[] such that if all elements of this sub-array are represented in binary notation and concatenated to form a binary string then number of 1’s in the resulting string is at least k. If no such sub-array exists then print -1.
Examples:
Input: arr[] = {4, 3, 7, 9}, k = 4
Output: 2
A possible sub-array is {3, 7}.Input: arr[] = {1, 2, 4, 8}, k = 2
Output: 2
Approach: The idea is to use two variables j and i and initialize them to 0 and 1 respectively, and an array count_one which will store the number of one’s present in the binary representation of a particular element of the array and a variable sum to store the number of 1’s upto ith index and ans to store the minimum length. Now iterate over the array, if the number of 1’s of ith or jth element of count_one is equal to k, then update ans as 1, if the sum of number of 1’s upto ith element is greater than or equal to k update ans as minimum of ans and (i-j)+1, else if it is less than k then increment j by 1, to increase the value of sum.
Below is the implementation of the approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Finds the sub-array with maximum lengthint FindSubarray(int arr[], int n, int k){ // Array which stores number of ones // present in the binary representation // of ith element of the array int count_one[n]; for (int i = 0; i < n; i++) { count_one[i] = __builtin_popcount(arr[i]); } // Sum variable to store sum of // number of ones // Initialize it as number of ones // present in the binary representation // of 0th element of the array int sum = count_one[0]; // If there is only a single element if (n == 1) { if (count_one[0] >= k) return 1; else return -1; } // Stores the minimum length // of the required sub-array int ans = INT_MAX; int i = 1; int j = 0; while (i < n) { // If binary representation of jth // element of array has 1's equal to k if (k == count_one[j]) { ans = 1; break; } // If binary representation of ith // element of array has 1's equal to k else if (k == count_one[i]) { ans = 1; break; } // If sum of number of 1's of // binary representation of elements upto // ith element is less than k else if (sum + count_one[i] < k) { sum += count_one[i]; i++; } // If sum of number of 1's of // binary representation of elements upto // ith element is greater than k else if (sum + count_one[i] > k) { ans = min(ans, (i - j) + 1); sum -= count_one[j]; j++; } else if (sum + count_one[i] == k) { ans = min(ans, (i - j) + 1); sum += count_one[i]; i++; } } if (ans != INT_MAX) return ans; else return -1;}// Driver codeint main(){ int arr[] = { 1, 2, 4, 8 }; int n = sizeof(arr) / sizeof(int); int k = 2; cout << FindSubarray(arr, n, k); return 0;} |
Java
// Java implementation of the approachclass GFG{// Finds the sub-array with maximum lengthstatic int FindSubarray(int arr[], int n, int k){ // Array which stores number of ones // present in the binary representation // of ith element of the array int []count_one = new int[n]; for (int i = 0; i < n; i++) { count_one[i] = Integer.bitCount(arr[i]); } // Sum variable to store sum of // number of ones // Initialize it as number of ones // present in the binary representation // of 0th element of the array int sum = count_one[0]; // If there is only a single element if (n == 1) { if (count_one[0] >= k) return 1; else return -1; } // Stores the minimum length // of the required sub-array int ans = Integer.MAX_VALUE; int i = 1; int j = 0; while (i < n) { // If binary representation of jth // element of array has 1's equal to k if (k == count_one[j]) { ans = 1; break; } // If binary representation of ith // element of array has 1's equal to k else if (k == count_one[i]) { ans = 1; break; } // If sum of number of 1's of // binary representation of elements upto // ith element is less than k else if (sum + count_one[i] < k) { sum += count_one[i]; i++; } // If sum of number of 1's of // binary representation of elements upto // ith element is greater than k else if (sum + count_one[i] > k) { ans = Math.min(ans, (i - j) + 1); sum -= count_one[j]; j++; } else if (sum + count_one[i] == k) { ans = Math.min(ans, (i - j) + 1); sum += count_one[i]; i++; } } if (ans != Integer.MAX_VALUE) return ans; else return -1;}// Driver codepublic static void main(String[] args) { int arr[] = { 1, 2, 4, 8 }; int n = arr.length; int k = 2; System.out.println(FindSubarray(arr, n, k));}}// This code is contributed by Princi Singh |
Python3
# Python3 implementation of the approach import sys;# Finds the sub-array with maximum length def FindSubarray(arr, n, k) : # Array which stores number of ones # present in the binary representation # of ith element of the array count_one = [0] * n; for i in range(n) : count_one[i] = bin(arr[i]).count('1'); # Sum variable to store sum of # number of ones # Initialize it as number of ones # present in the binary representation # of 0th element of the array sum = count_one[0]; # If there is only a single element if (n == 1) : if (count_one[0] >= k) : return 1; else : return -1; # Stores the minimum length # of the required sub-array ans = sys.maxsize; i = 1; j = 0; while (i < n) : # If binary representation of jth # element of array has 1's equal to k if (k == count_one[j]) : ans = 1; break; # If binary representation of ith # element of array has 1's equal to k elif (k == count_one[i]) : ans = 1; break; # If sum of number of 1's of # binary representation of elements upto # ith element is less than k elif (sum + count_one[i] < k) : sum += count_one[i]; i += 1; # If sum of number of 1's of # binary representation of elements upto # ith element is greater than k elif (sum + count_one[i] > k) : ans = min(ans, (i - j) + 1); sum -= count_one[j]; j += 1; elif (sum + count_one[i] == k) : ans = min(ans, (i - j) + 1); sum += count_one[i]; i += 1; if (ans != sys.maxsize) : return ans; else : return -1; # Driver code if __name__ == "__main__" : arr = [ 1, 2, 4, 8 ]; n = len(arr); k = 2; print(FindSubarray(arr, n, k)); # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System; class GFG{// Finds the sub-array with maximum lengthstatic int FindSubarray(int []arr, int n, int k){ // Array which stores number of ones // present in the binary representation // of ith element of the array int []count_one = new int[n]; int i = 0; for (i = 0; i < n; i++) { count_one[i] = bitCount(arr[i]); } // Sum variable to store sum of // number of ones // Initialize it as number of ones // present in the binary representation // of 0th element of the array int sum = count_one[0]; // If there is only a single element if (n == 1) { if (count_one[0] >= k) return 1; else return -1; } // Stores the minimum length // of the required sub-array int ans = int.MaxValue; i = 1; int j = 0; while (i < n) { // If binary representation of jth // element of array has 1's equal to k if (k == count_one[j]) { ans = 1; break; } // If binary representation of ith // element of array has 1's equal to k else if (k == count_one[i]) { ans = 1; break; } // If sum of number of 1's of // binary representation of elements upto // ith element is less than k else if (sum + count_one[i] < k) { sum += count_one[i]; i++; } // If sum of number of 1's of // binary representation of elements upto // ith element is greater than k else if (sum + count_one[i] > k) { ans = Math.Min(ans, (i - j) + 1); sum -= count_one[j]; j++; } else if (sum + count_one[i] == k) { ans = Math.Min(ans, (i - j) + 1); sum += count_one[i]; i++; } } if (ans != int.MaxValue) return ans; else return -1;}static int bitCount(long x){ int setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits;}// Driver codepublic static void Main(String[] args) { int []arr = { 1, 2, 4, 8 }; int n = arr.Length; int k = 2; Console.WriteLine(FindSubarray(arr, n, k));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation of the approach// Finds the sub-array with maximum lengthfunction FindSubarray(arr, n, k){ // Array which stores number of ones // present in the binary representation // of ith element of the array let count_one = new Array(n); let i = 0; for(i = 0; i < n; i++) { count_one[i] = bitCount(arr[i]); } // Sum variable to store sum of // number of ones // Initialize it as number of ones // present in the binary representation // of 0th element of the array let sum = count_one[0]; // If there is only a single element if (n == 1) { if (count_one[0] >= k) return 1; else return -1; } // Stores the minimum length // of the required sub-array let ans = Number.MAX_VALUE; i = 1; let j = 0; while (i < n) { // If binary representation of jth // element of array has 1's equal to k if (k == count_one[j]) { ans = 1; break; } // If binary representation of ith // element of array has 1's equal to k else if (k == count_one[i]) { ans = 1; break; } // If sum of number of 1's of // binary representation of elements upto // ith element is less than k else if (sum + count_one[i] < k) { sum += count_one[i]; i++; } // If sum of number of 1's of // binary representation of elements upto // ith element is greater than k else if (sum + count_one[i] > k) { ans = Math.min(ans, (i - j) + 1); sum -= count_one[j]; j++; } else if (sum + count_one[i] == k) { ans = Math.min(ans, (i - j) + 1); sum += count_one[i]; i++; } } if (ans != Number.MAX_VALUE) return ans; else return -1;}function bitCount(x){ let setBits = 0; while (x != 0) { x = x & (x - 1); setBits++; } return setBits;}// Driver codelet arr = [ 1, 2, 4, 8 ];let n = arr.length;let k = 2;document.write(FindSubarray(arr, n, k));// This code is contributed by divyeshrabadiya07</script> |
Output:
2
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