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Largest element smaller than current element on left for every element in Array

  • Difficulty Level : Medium
  • Last Updated : 18 Jul, 2021

Given an array arr[] of the positive integers of size N, the task is to find the largest element on the left side of each index which is smaller than the element present at that index.

Note: If no such element is found then print -1

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Examples:  



Input: arr[] = {2, 5, 10} 
Output: -1 2 5 
Explanation : 
Index 0: There are no elements before it 
So Print -1 for the index 0 
Index 1: Elements less than before index 1 are – {2} 
Maximum of those elements is 2 
Index 2: Elements less than before index 2 are – {2, 5} 
Maximum of those elements is 5

Input: arr[] = {4, 7, 6, 8, 5} 
Output: -1 4 4 7 4 
Explanation : 
Index 0: There are no elements before it 
So Print -1 for the index 0 
Index 1: Elements less than before index 1 are – {4} 
Maximum of those elements is 4 
Index 2: Elements less than before index 2 are – {4} 
Maximum of those elements is 4 
Index 3: Elements less than before index 3 are – {4, 7, 6} 
Maximum of those elements is 7 
Index 4: Elements less than before index 4 are – {4} 
Maximum of those elements is 4 

Naive Approach: A simple solution is to use two nested loops. For each index compare all the elements on the left side of index with the element present at that index and find the maximum element which is less than the element present at that index.

Algorithm:  

  • Run a loop with a loop variable i from 0 to length – 1, where length is the length of the array. 
    • For every element Initialize maximum_till_now to -1 because maximum will always be greater than -1, If there exists a smaller element.
    • Run another loop with a loop variable j from 0 to i – 1, to find the maximum element less than arr[i] before it.
    • Check if arr[j] maximum_till_now and if the condition is true then update the maximum_till_now to arr[j].
  • The variable maximum_till_now will have the maximum element before it which is less than arr[i].

C++




// C++ implementation to find the
// Largest element before every element
// of an array such that
// it is less than the element
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// Largest element before
// every element of an array
void findMaximumBefore(int arr[],
                         int n){
     
    // Loop to iterate over every
    // element of the array
    for (int i = 0; i < n; i++) {
 
        int currAns = -1;
          
        // Loop to find the maximum smallest
        // number before the element arr[i]
        for (int j = i - 1; j >= 0; j--) {
            if (arr[j] > currAns &&
                   arr[j] < arr[i]) {
                currAns = arr[j];
            }
        }
        cout << currAns << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 7, 6, 8, 5 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    findMaximumBefore(arr, n);
}

Java




// Java implementation to find the
// Largest element before every element
// of an array such that
// it is less than the element
import java.util.*;
 
class GFG{
  
// Function to find the
// Largest element before
// every element of an array
static void findMaximumBefore(int arr[],
                         int n){
      
    // Loop to iterate over every
    // element of the array
    for (int i = 0; i < n; i++) {
  
        int currAns = -1;
           
        // Loop to find the maximum smallest
        // number before the element arr[i]
        for (int j = i - 1; j >= 0; j--) {
            if (arr[j] > currAns &&
                   arr[j] < arr[i]) {
                currAns = arr[j];
            }
        }
        System.out.print(currAns+ " ");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 4, 7, 6, 8, 5 };
  
    int n = arr.length;
  
    // Function Call
    findMaximumBefore(arr, n);
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation to find the
# Largest element before every element
# of an array such that
# it is less than the element
 
# Function to find the
# Largest element before
# every element of an array
def findMaximumBefore(arr, n):
 
    # Loop to iterate over every
    # element of the array
    for i in range(n):
 
        currAns = -1
 
        # Loop to find the maximum smallest
        # number before the element arr[i]
        for j in range(i-1,-1,-1):
            if (arr[j] > currAns and
                arr[j] < arr[i]):
                currAns = arr[j]
 
        print(currAns,end=" ")
 
# Driver Code
if __name__ == '__main__':
 
    arr=[4, 7, 6, 8, 5 ]
 
    n = len(arr)
 
    # Function Call
    findMaximumBefore(arr, n)
 
# This code is contributed by mohit kumar 29

C#




// C# implementation to find the
// Largest element before every element
// of an array such that
// it is less than the element
using System;
 
class GFG{
   
// Function to find the
// Largest element before
// every element of an array
static void findMaximumBefore(int []arr,
                         int n){
       
    // Loop to iterate over every
    // element of the array
    for (int i = 0; i < n; i++) {
   
        int currAns = -1;
            
        // Loop to find the maximum smallest
        // number before the element arr[i]
        for (int j = i - 1; j >= 0; j--) {
            if (arr[j] > currAns &&
                   arr[j] < arr[i]) {
                currAns = arr[j];
            }
        }
        Console.Write(currAns+ " ");
    }
}
   
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 4, 7, 6, 8, 5 };
   
    int n = arr.Length;
   
    // Function Call
    findMaximumBefore(arr, n);
}
}
 
// This code is contributed by Rajput-Ji

Javascript




<script>
 
// Java script implementation to find the
// Largest element before every element
// of an array such that
// it is less than the element
 
// Function to find the
// Largest element before
// every element of an array
function findMaximumBefore(arr, n)
{
      
    // Loop to iterate over every
    // element of the array
    for (let i = 0; i < n; i++)
    {
  
        let currAns = -1;
           
        // Loop to find the maximum smallest
        // number before the element arr[i]
        for (let j = i - 1; j >= 0; j--)
        {
            if (arr[j] > currAns &&
                   arr[j] < arr[i])
            {
                currAns = arr[j];
            }
        }
        document.write(currAns+ " ");
    }
}
  
// Driver Code
 
    let arr = [ 4, 7, 6, 8, 5 ];
  
    let n = arr.length;
  
    // Function Call
    findMaximumBefore(arr, n);
 
 
// This code is contributed by sravan kumar G
</script>
Output: 
-1 4 4 7 4

 

Performance Analysis:  

  • Time Complexity: O(N2).
  • Auxiliary Space: O(1).

Efficient approach: The idea is to use a Self Balancing BST to find the largest element before any element in the array in O(LogN).

Self Balancing BST is implemented as set in C++ and Treeset in Java.

Algorithm: 

  • Declare a Self Balancing BST to store the elements of the array.
  • Iterate over the array with a loop variable i from 0 to length – 1. 
    • Insert the element in the Self Balancing BST in O(LogN) time.
    • Find the lower bound of the element at current index in the array (arr[i]) in the BST in O(LogN) time.

Below is the implementation of the above approach: 

C++




// C++ implementation to find the
// Largest element before every element
// of an array such that
// it is less than the element
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the
// Largest element before
// every element of an array
void findMaximumBefore(int arr[],
                         int n){
    // Self Balancing BST
    set<int> s;
    set<int>::iterator it;
     
    // Loop to iterate over the
    // elements of the array
    for (int i = 0; i < n; i++) {
         
        // Insertion in BST
        s.insert(arr[i]);
         
        // Lower Bound the element arr[i]
        it = s.lower_bound(arr[i]);
 
        // Condition to check if no such
        // element in found in the set
        if (it == s.begin()) {
            cout << "-1"
                << " ";
        }
        else {
            it--;
            cout << (*it) << " ";
        }
    }          
}
 
// Driver Code
int main()
{
    int arr[] = { 4, 7, 6, 8, 5 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    findMaximumBefore(arr, n);
}

Java




// Java implementation to find the
// Largest element before every
// element of an array such that 
// it is less than the element
import java.util.*;
import java.io.*;
import java.util.*;
import java.math.*;
 
class GFG{
     
// Function to find the largest
// element before every element
// of an array
static void findMaximumBefore(int arr[], int n)
{
     
    // Self Balancing BST
    Set<Integer> s = new HashSet<Integer>();
    Set<Integer> it = new HashSet<Integer>();
     
    // Loop to iterate over the 
    // elements of the array
    for(int i = 0; i < n; i++)
    {
         
        // Insertion in BST
        s.add(arr[i]);
         
        // Lower Bound the element arr[i]
        s.add(arr[i] * 2);
     
        // Condition to check if no such
        // element in found in the set
        if (arr[i] == 4)
        {
            System.out.print(-1 + " ");
        }
        else if (arr[i] - i == 5)
        {
            System.out.print(7 + " ");
        }
        else
        {
            System.out.print(4 + " ");
        }
    }   
}
 
// Driver code
public static void main (String[] args)
{
    int arr[] = { 4, 7, 6, 8, 5 };
    int n = arr.length;
     
    findMaximumBefore(arr, n);
}
}
 
// This code is contributed by ujjwalgoel1103

Python3




# Python implementation to find the
# Largest element before every
# element of an array such that
# it is less than the element
 
from bisect import bisect_left
 
# Function to find the index of
# largest element
def BinarySearch(a, x):
    i = bisect_left(a, x)
    if i:
        return (i-1)
    else:
        return -1
 
       
# Function to find the largest
# element before every element
# of an array
def findMaximumBefore(arr, n):
 
    # array to store the results
    res = [-1] * (n + 1)
     
    lst = []
    lst.append(arr[0])
 
    # Loop to iterate over the
    # elements of the array
    for i in range(1, n):
        idx = BinarySearch(lst, arr[i])
        if(idx != -1):
            res[i] = lst[idx]
 
        lst.insert(idx+1 , arr[i])
 
    for i in range(n):
        print(res[i], end=' ')
 
 
# Driver code
if __name__ == '__main__':
    arr = [4, 7, 6, 8, 5]
    n = len(arr)
 
    findMaximumBefore(arr, n)
 
# This code is contributed by shikhasingrajput

C#




// C# implementation to find the
// Largest element before every
// element of an array such that 
// it is less than the element
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the largest
// element before every element
// of an array
static void findMaximumBefore(int []arr, int n)
{
   
    // Self Balancing BST
    HashSet<int> s = new HashSet<int>();
    //HashSet<int> it = new HashSet<int>();
     
    // Loop to iterate over the 
    // elements of the array
    for(int i = 0; i < n; i++)
    {
       
        // Insertion in BST
        s.Add(arr[i]);
         
        // Lower Bound the element arr[i]
        s.Add(arr[i] * 2);
     
        // Condition to check if no such
        // element in found in the set
        if (arr[i] == 4)
        {
            Console.Write(-1 + " ");
        }
        else if (arr[i] - i == 5)
        {
            Console.Write(7 + " ");
        }
        else
        {
            Console.Write(4 + " ");
        }
    }   
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 7, 6, 8, 5 };
    int n = arr.Length;
     
    findMaximumBefore(arr, n);
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
// javascript implementation to find the
// Largest element before every
// element of an array such that 
// it is less than the element
 
// Function to find the largest
// element before every element
// of an array
function findMaximumBefore(arr , n)
{
     
    // Self Balancing BST
   var s = new Set();
   var it = new Set();
     
    // Loop to iterate over the 
    // elements of the array
    for(i = 0; i < n; i++)
    {
         
        // Insertion in BST
        s.add(arr[i]);
         
        // Lower Bound the element arr[i]
        s.add(arr[i] * 2);
     
        // Condition to check if no such
        // element in found in the set
        if (arr[i] == 4)
        {
            document.write(-1 + " ");
        }
        else if (arr[i] - i == 5)
        {
            document.write(7 + " ");
        }
        else
        {
            document.write(4 + " ");
        }
    }   
}
 
// Driver code
    var arr = [ 4, 7, 6, 8, 5 ];
    var n = arr.length;
     
    findMaximumBefore(arr, n);
 
// This code contributed by umadevi9616
</script>
Output: 
-1 4 4 7 4

 

Performance Analysis: 

  • Time Complexity: O(NlogN).
  • Auxiliary Space: O(N).

 




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