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Length of second longest sequence of consecutive 1s in a binary array
  • Difficulty Level : Basic
  • Last Updated : 16 Apr, 2021

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: 



  1. Initialise variables maxi, count and second_max to store the length of the longest, current and second longest sequence of consecutive 1s respectively..
  2. Iterate over the given array twice.
  3. 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.
  4. 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 length
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 = 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 Code
int 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 approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to find maximum
// and second maximum length
static 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 length
def 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 Code
if __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 approach
using System.Collections.Generic;
using System;
 
class GFG{
     
// Function to find maximum
// and second maximum length
static 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 length
function 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);
  
    // Prlet 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)
 

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