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Longest subarray with only one value greater than k

  • Difficulty Level : Hard
  • Last Updated : 25 May, 2021

Given an array of N numbers, find length of the longest subarray such that K is the second largest element on insertion. 
Examples: 
 

Input: a[] = {9, 5, 5, 6, 8}, K = 7 
Output:
The longest subarray is {9, 5, 5, 6}, in which if K is inserted it becomes {9, 5, 5, 6, 7}, and 
7 is the second largest element in the array 
Input: a[] = {9, 5, 5, 6, 8}, K = 10 
Output:
Since the maximum number in the array is less than that of K, hence it is not possible. 
Input: a[] = {8, 5, 10, 10, 8}, K = 9 
Output:
9 is the second largest element of whole array 
 

 

A naive approach is to iterate for every possible subarray and check if on insertion of K, it becomes the second largest element or not. Naively we can store the length of the longest of all such subarrays possible. 
Time Complexity: O(N^2) 
An efficient solution is to use the two pointer technique to solve the above problem. Below is the algorithm to solve the above problem. 
 

  • Initialize two pointers front and end as 0, and a visited array which marks the index has been visited or not.
  • We need a set container so that we can the second largest element in any range front-end in O(log N) and a unordered_map to count the frequencies of the elements in the array to decide on the deletion from the set.
  • Initially check if an element exists or not which is greater than K or not, if there is no such element, then subarray is not possible.
  • Insert the element at a[end] in the set and increase its frequency in the map if the index end has not been visited before to avoid multiple insertions of same index.
  • If the set contains only one element then insertion of K is possible as we have only elements and since we know there exists at least one element > k, then this subarray can be a part of the longest subarray, hence we count it and move forward the end pointer.
  • If the set contains more then one element, then the s.end() pointer points after the last element, hence decreasing it twice will give us our second largest element in the range front-end.
  • If the second largest element is greater than K, then this subarray is not possible, hence we need to move the first pointer ahead, but before doing this, check if the frequency of a[front] is 1 or not, if it is so, then delete it from the set, else just decrease the frequency in map by 1, since the element will be existing in any index>front. Increase the front pointer by one.
  • If the second largest element is not greater than K, then simply increase the end pointer by one.
  • Store the largest of lengths of end-front and return it.

Below is the implementation of the above approach: 
 



C++




// C++ program to find the length of the longest
// subarray such that K is the second largest element
// on insertion
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of longest subarray
int lengthOfLongestSubarray(int a[], int n, int k)
{
    bool flag = 0;
 
    // Check if any element exists which is
    // greater than k or not
    for (int i = 0; i < n; i++) {
        if (a[i] > k)
            flag = 1;
    }
 
    if (!flag) {
        return 0;
    }
 
    // two pointers used
    int front = 0;
    int end = 0;
 
    // find the maximum length
    int maxi = 0;
 
    // Map used to count frequencies
    unordered_map<int, int> mpp;
 
    // set used to find the second largest
    // element in range front-end
    set<int> s;
 
    // initialize all index of array as 0
    bool vis[n];
    memset(vis, 0, sizeof vis);
 
    // iterate till any of the pointer exceeds N
    while (front < n && end < n) {
 
        // length of longest subarray
        maxi = max(maxi, end - front);
 
        // if the current index has not been
        // visited previously then insert it
        // in the set and increase the frequency
        // and mark it as visited.
        if (!vis[end]) {
            mpp[a[end]]++;
            s.insert(a[end]);
            vis[end] = 1;
        }
 
        // find the largest element in the set
        auto it = s.end();
 
        // if only one element is there in set,
        // then insertion of K is possible which
        // will include other elements
        if (s.size() == 1) {
 
            // increase the second pointer in order
            // to include more elements in the subarray
            end++;
            continue;
        }
 
        // twice decrease the
        // iterator as s.end() points to
        // after the last element
        it--;
 
        // second largest element in set
        it--;
        int el = *it;
 
        // if the second largest element is greater than the
        // K, then it is not possible to insert element
        // in range front-end, and thus decrease the
        // frequency of a[front] and remove from set
        // accordingly
        if (el > k) {
            if (mpp[a[front]] == 1) {
                s.erase(a[front]);
                mpp[a[front]]--;
            }
            else
                mpp[a[front]]--;
 
            // move ahead the first pointer
            front++;
        }
        else {
 
            // increase the second pointer
            // if the second largest element is smaller
            // than or equals to K
            end++;
        }
    }
 
    // at then end also check for last subarray length
    maxi = max(maxi, end - front);
 
    return maxi;
}
 
// Driver Code
int main()
{
    int a[] = { 9, 5, 5, 6, 8 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 7;
 
    cout << lengthOfLongestSubarray(a, n, k);
 
    return 0;
}

Java




// Java program to find the length of the longest
// subarray such that K is the second largest element
// on insertion
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to find the length of longest subarray
  static int lengthOfLongestSubarray(int a[], int n, int k)
  {
    int flag = 0;
 
    // Check if any element exists which is
    // greater than k or not
    for (int i = 0; i < n; i++)
    {
      if (a[i] > k)
        flag = 1;
    }
    if (flag == 0)
    {
      return 0;
    }
 
    // two pointers used
    int front = 0;
    int end = 0;
 
    // find the maximum length
    int maxi = 0;    
    Map<Integer, Integer> mpp = new HashMap<Integer, Integer>();
    Set<Integer> s = new HashSet<Integer>();
 
    // initialize all index of array as 0
    int[] vis = new int[n];
 
    // iterate till any of the pointer exceeds N
    while (front < n && end < n)
    {
 
      // length of longest subarray
      maxi = Math.max(maxi, end - front);
 
      // if the current index has not been
      // visited previously then insert it
      // in the set and increase the frequency
      // and mark it as visited.        
      if (vis[end] == 0)
      {
        if(mpp.containsKey(a[end]))
        {
          mpp.put(a[end], mpp.get(a[end]) + 1);
        }
        else
        {
          mpp.put(a[end], 1);
        }
        s.add(a[end]);
        vis[end] = 1;
      }     
      int it = s.size();
      List<Integer> S = new ArrayList<Integer>(s);
      Collections.sort(S);
 
      // if only one element is there in set,
      // then insertion of K is possible which
      // will include other elements
      if (S.size() == 1)
      {
 
        // increase the second pointer in order
        // to include more elements in the subarray
        end++;
        continue;
      }
 
      // twice decrease the
      // iterator as s.end() points to
      // after the last element
      it--;
 
      // second largest element in set
      it--;
      int el = S.get(it);
 
      // if the second largest element is greater than the
      // K, then it is not possible to insert element
      // in range front-end, and thus decrease the
      // frequency of a[front] and remove from set
      // accordingly
 
      if (el > k)
      {
        if(mpp.get(a[front]) == 1)
        {
          mpp.put(a[front], mpp.get(a[front]) - 1);
        }
        else
        {
          mpp.put(a[front], mpp.get(a[front]) - 1);  
        }
 
        // move ahead the first pointer
        front++;
      }
      else
      {
 
        // increase the second pointer
        // if the second largest element is smaller
        // than or equals to K
        end++;
      }
    }
 
    // at then end also check for last subarray length
    maxi = Math.max(maxi, end - front);
    return maxi;
 
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int[] a = { 9, 5, 5, 6, 8 };
    int n = a.length;
    int k = 7;
    System.out.println(lengthOfLongestSubarray(a, n, k));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 program to find the length of
# the longest subarray such that K is
# the second largest element on insertion
 
# Function to find the length of longest subarray
def lengthOfLongestSubarray(a, n, k):
    flag = 0
 
    # Check if any element exists which is
    # greater than k or not
    for i in range(n):
        if (a[i] > k):
            flag = 1
 
    if (flag == 0):
        return 0
 
    # two pointers used
    front = 0
    end = 0
 
    # find the maximum length
    maxi = 0
 
    # Map used to count frequencies
    mpp = dict()
 
    # set used to find the second largest
    # element in range front-end
    s = dict()
 
    # initialize all index of array as 0
    vis = [0] * n
 
    # iterate till any of the pointer exceeds N
    while (front < n and end < n):
 
        # length of longest subarray
        maxi = max(maxi, end - front)
 
        # if the current index has not been
        # visited previously then insert it
        # in the set and increase the frequency
        # and mark it as visited.
        if (vis[end] == 0):
            mpp[a[end]] = mpp.get(a[end], 0) + 1
            s[a[end]] = s.get(a[end], 0) + 1
            vis[end] = 1
 
        # find the largest element in the set
        iit = sorted(list(s))
        it = len(iit)
         
        # if only one element is there in set,
        # then insertion of K is possible which
        # will include other elements
        if (len(s) == 1):
 
            # increase the second pointer in order
            # to include more elements in the subarray
            end += 1
            continue
 
        # twice decrease the
        # iterator as s.end() points to
        # after the last element
        it -= 1
 
        # second largest element in set
        it -= 1
        el = iit[it]
 
        # if the second largest element is greater than the
        # K, then it is not possible to insert element
        # in range front-end, and thus decrease the
        # frequency of a[front] and remove from set
        # accordingly
        if (el > k):
            if (mpp[a[front]] == 1):
                del s[a[front]]
                mpp[a[front]] -= 1
            else:
                mpp[a[front]] -= 1
 
            # move ahead the first pointer
            front += 1
        else :
 
            # increase the second pointer
            # if the second largest element is
            # smaller than or equals to K
            end += 1
 
    # at then end also check for
    # last subarray length
    maxi = max(maxi, end - front)
 
    return maxi
 
# Driver Code
= [9, 5, 5, 6, 8]
n = len(a)
k = 7
 
print(lengthOfLongestSubarray(a, n, k))
 
# This code is contributed by Mohit Kumar

C#




// C# program to find the length of the longest
// subarray such that K is the second largest element
// on insertion
using System;
using System.Collections.Generic;
public class GFG
{
 
  // Function to find the length of longest subarray
  static int lengthOfLongestSubarray(int[] a, int n, int k)
  {
    int flag = 0;
 
    // Check if any element exists which is
    // greater than k or not
    for (int i = 0; i < n; i++)
    {
      if (a[i] > k)
        flag = 1;
    }
    if (flag == 0)
    {
      return 0;
    }
 
    // two pointers used
    int front = 0;
    int end = 0;
 
    // find the maximum length
    int maxi = 0;    
 
    Dictionary<int, int> mpp =new Dictionary<int, int>();
    SortedSet<int> s = new SortedSet<int>();
 
    // initialize all index of array as 0
    int[] vis = new int[n];
 
    // iterate till any of the pointer exceeds N
    while (front < n && end < n)
    {
 
      // length of longest subarray
      maxi = Math.Max(maxi, end - front);  
 
      // if the current index has not been
      // visited previously then insert it
      // in the set and increase the frequency
      // and mark it as visited.        
      if (vis[end] == 0)
      {
        if(mpp.ContainsKey(a[end]))
        {
          mpp[a[end]]++;
        }
        else
        {
          mpp.Add(a[end],1);
        }
        s.Add(a[end]);
        vis[end] = 1;
      }
      int it = s.Count;
      List<int> S = new List<int>(s);
      if(S.Count == 1)
      {
        // increase the second pointer in order
        // to include more elements in the subarray
        end++;
        continue;
      }
 
      // twice decrease the
      // iterator as s.end() points to
      // after the last element
      it--;
 
      // second largest element in set
      it--;
      int el = S[it];
 
      // if the second largest element is greater than the
      // K, then it is not possible to insert element
      // in range front-end, and thus decrease the
      // frequency of a[front] and remove from set
      // accordingly        
      if (el > k)
      {
        if(mpp[a[front]] == 1)
        {
          mpp[a[front]]--;
        }
        else
        {
          mpp[a[front]]--;
        }
        front++;
      }
      else
      {
        // increase the second pointer
        // if the second largest element is smaller
        // than or equals to K
        end++;
      }
    }
 
    // at then end also check for last subarray length
    maxi = Math.Max(maxi, end - front);
    return maxi;
  }
 
  // Driver code
  static public void Main (){
    int[] a = { 9, 5, 5, 6, 8 };
    int n = a.Length;
    int k = 7;
 
    Console.WriteLine(lengthOfLongestSubarray(a, n, k));
  }
}
 
// This code is contributed by rag2127

Javascript




<script>
// Javascript program to find the length of the longest
// subarray such that K is the second largest element
// on insertion
     
    // Function to find the length of longest subarray
    function lengthOfLongestSubarray(a,n,k)
    {
        let flag = 0;
  
    // Check if any element exists which is
    // greater than k or not
    for (let i = 0; i < n; i++)
    {
      if (a[i] > k)
        flag = 1;
    }
    if (flag == 0)
    {
      return 0;
    }
  
    // two pointers used
    let front = 0;
    let end = 0;
  
    // find the maximum length
    let maxi = 0;   
    let mpp = new Map();
    let s = new Set();
  
    // initialize all index of array as 0
    let vis = new Array(n);
     for(let i=0;i<n;i++)
    {
        vis[i]=0;
    }
    // iterate till any of the pointer exceeds N
    while (front < n && end < n)
    {
  
      // length of longest subarray
      maxi = Math.max(maxi, end - front);
  
      // if the current index has not been
      // visited previously then insert it
      // in the set and increase the frequency
      // and mark it as visited.       
      if (vis[end] == 0)
      {
        if(mpp.has(a[end]))
        {
          mpp.set(a[end], mpp.get(a[end]) + 1);
        }
        else
        {
          mpp.set(a[end], 1);
        }
        s.add(a[end]);
        vis[end] = 1;
      }    
      let it = s.size;
      let S = Array.from(s);
      S.sort(function(a,b){return a-b;});
  
      // if only one element is there in set,
      // then insertion of K is possible which
      // will include other elements
      if (S.length == 1)
      {
  
        // increase the second pointer in order
        // to include more elements in the subarray
        end++;
        continue;
      }
  
      // twice decrease the
      // iterator as s.end() points to
      // after the last element
      it--;
  
      // second largest element in set
      it--;
      let el = S[it];
  
      // if the second largest element is greater than the
      // K, then it is not possible to insert element
      // in range front-end, and thus decrease the
      // frequency of a[front] and remove from set
      // accordingly
  
      if (el > k)
      {
        if(mpp.get(a[front]) == 1)
        {
          mpp.set(a[front], mpp.get(a[front]) - 1);
        }
        else
        {
          mpp.set(a[front], mpp.get(a[front]) - 1); 
        }
  
        // move ahead the first pointer
        front++;
      }
      else
      {
  
        // increase the second pointer
        // if the second largest element is smaller
        // than or equals to K
        end++;
      }
    }
  
    // at then end also check for last subarray length
    maxi = Math.max(maxi, end - front);
    return maxi;
    }
     
     // Driver Code
    let a=[ 9, 5, 5, 6, 8 ];
    let n = a.length;
    let k = 7;
    document.write(lengthOfLongestSubarray(a, n, k));
     
     
// This code is contributed by patel2127
</script>
Output: 
4

 

Time Complexity: O(N * log N)
 

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