Given a binary array arr[] of size N, the task is to find the length of the second longest sequence of consecutive 1s present in the array.
Examples:
Input: arr[] = {1, 1, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0}
Output: 4 3
Explanation:
Longest sequence of consecutive ones is 4 i.e {arr[7], … arr[10]}.
Second longest sequence of consecutive ones is 3 i.e {arr[0], … arr[2]}.Input: arr[] = {1, 0, 1}
Output: 1 0
Approach: The idea is to traverse the given binary array and keep track of the largest and the second largest length of consecutive one’s encountered so far. Below are the steps:
- Initialise variables maxi, count and second_max to store the length of the longest, current and second longest sequence of consecutive 1s respectively..
- Iterate over the given array twice.
- First, traverse the array from left to right. For every 1 encountered, then increment count and compare it with maximum so far. If 0 is encountered, reset count as 0.
- In the second traversal, find the required second longest count of consecutive 1’s following the above procedure.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find maximum// and second maximum lengthvoid FindMax(int arr[], int N){ // Initialise maximum length int maxi = -1; // Initialise second maximum length int maxi2 = -1; // Initialise count int count = 0; // Iterate over the array for (int i = 0; i < N; ++i) { // If sequence ends if (arr[i] == 0) // Reset count count = 0; // Otherwise else { // Increase length // of current sequence count++; // Update maximum maxi = max(maxi, count); } } // Traverse the given array for (int i = 0; i < N; i++) { // If sequence continues if (arr[i] == 1) { // Increase length // of current sequence count++; // Update second max if (count > maxi2 && count < maxi) { maxi2 = count; } } // Reset count when 0 is found if (arr[i] == 0) count = 0; } maxi = max(maxi, 0); maxi2 = max(maxi2, 0); // Print the result cout << maxi2;}// Driver Codeint main(){ // Given array int arr[] = { 1, 1, 1, 0, 0, 1, 1 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call FindMax(arr, N); return 0;} |
Java
// Java implementation of the above approachimport java.io.*; import java.util.*; class GFG{ // Function to find maximum// and second maximum lengthstatic void FindMax(int arr[], int N){ // Initialise maximum length int maxi = -1; // Initialise second maximum length int maxi2 = -1; // Initialise count int count = 0; // Iterate over the array for(int i = 0; i < N; ++i) { // If sequence ends if (arr[i] == 0) // Reset count count = 0; // Otherwise else { // Increase length // of current sequence count++; // Update maximum maxi = Math.max(maxi, count); } } // Traverse the given array for(int i = 0; i < N; i++) { // If sequence continues if (arr[i] == 1) { // Increase length // of current sequence count++; // Update second max if (count > maxi2 && count < maxi) { maxi2 = count; } } // Reset count when 0 is found if (arr[i] == 0) count = 0; } maxi = Math.max(maxi, 0); maxi2 = Math.max(maxi2, 0); // Print the result System.out.println( maxi2);}// Driver code public static void main(String[] args) { int arr[] = { 1, 1, 1, 0, 0, 1, 1 }; int n = arr.length; FindMax(arr, n);} } // This code is contributed by Stream_Cipher |
Python3
# Python3 implementation of the above approach# Function to find maximum# and second maximum lengthdef FindMax(arr, N): # Initialise maximum length maxi = -1 # Initialise second maximum length maxi2 = -1 # Initialise count count = 0 # Iterate over the array for i in range(N): # If sequence ends if (arr[i] == 0): # Reset count count = 0 # Otherwise else: # Increase length # of current sequence count += 1 # Update maximum maxi = max(maxi, count) # Traverse the given array for i in range(N): # If sequence continues if (arr[i] == 1): # Increase length # of current sequence count += 1 # Update second max if (count > maxi2 and count < maxi): maxi2 = count # Reset count when 0 is found if (arr[i] == 0): count = 0 maxi = max(maxi, 0) maxi2 = max(maxi2, 0) # Print the result print(maxi2)# Driver Codeif __name__ == '__main__': # Given array arr = [1, 1, 1, 0, 0, 1, 1] N = len(arr) # Function Call FindMax(arr, N)# This code is contributed by Mohit Kumar29 |
C#
// C# implementation of the above approachusing System.Collections.Generic; using System; class GFG{ // Function to find maximum// and second maximum lengthstatic void FindMax(int []arr, int N){ // Initialise maximum length int maxi = -1; // Initialise second maximum length int maxi2 = -1; // Initialise count int count = 0; // Iterate over the array for(int i = 0; i < N; ++i) { // If sequence ends if (arr[i] == 0) // Reset count count = 0; // Otherwise else { // Increase length // of current sequence count++; // Update maximum maxi = Math.Max(maxi, count); } } // Traverse the given array for(int i = 0; i < N; i++) { // If sequence continues if (arr[i] == 1) { // Increase length // of current sequence count++; // Update second max if (count > maxi2 && count < maxi) { maxi2 = count; } } // Reset count when 0 is found if (arr[i] == 0) count = 0; } maxi = Math.Max(maxi, 0); maxi2 = Math.Max(maxi2, 0); // Print the result Console.WriteLine( maxi2);}// Driver code public static void Main() { int []arr = { 1, 1, 1, 0, 0, 1, 1 }; int n = arr.Length; FindMax(arr, n);} } // This code is contributed by Stream_Cipher |
Javascript
<script>// JavaScript program to implement// the above approach// Function to find maximum// and second maximum lengthfunction FindMax(arr, N){ // Initialise maximum length let maxi = -1; // Initialise second maximum length let maxi2 = -1; // Initialise count let count = 0; // Iterate over the array for(let i = 0; i < N; ++i) { // If sequence ends if (arr[i] == 0) // Reset count count = 0; // Otherwise else { // Increase length // of current sequence count++; // Update maximum maxi = Math.max(maxi, count); } } // Traverse the given array for(let i = 0; i < N; i++) { // If sequence continues if (arr[i] == 1) { // Increase length // of current sequence count++; // Update second max if (count > maxi2 && count < maxi) { maxi2 = count; } } // Reset count when 0 is found if (arr[i] == 0) count = 0; } maxi = Math.max(maxi, 0); maxi2 = Math.max(maxi2, 0); // Print the result document.write( maxi2);}// Driver code let arr = [ 1, 1, 1, 0, 0, 1, 1 ]; let n = arr.length; FindMax(arr, n); // This code is contributed by target_2.</script> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
