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

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.

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 Intialize 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].

CPP

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// 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);
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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:

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// 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);
}

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Output:

-1 4 4 7 4 

Performance Analysis:

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

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