Closest greater or same value on left side for every element in array

Given an array of integers, find the closest (not considering distance, but value) greater or same value on left of every element. If an element has no greater or same value on left side, print -1.

Examples:

Input : arr[] = {10, 5, 11, 6, 20, 12}
Output : -1, 10, -1, 10, -1, 20
First element has nothing on left side, so answer for first is -1.
Second element 5 has 10 on left, so answer is 10.
Third element 11 has nothing greater or same, so answer is -1.
Fourth element 6 has 10 as value wise closes, so answer is 10
Similarly we get values for fifth and sixth elements.



Input : arr[] = {10, 5, 11, 10, 20, 12}
Output : -1, 10, -1, 10, -1, 20

A simple solution is to run two nested loops. We pick an outer element one by one. For every picked element, we traverse toward left of it and find the closest (value wise) greater element. Time complexity of this solution is O(n*n)

An efficient solution is to use Self Balancing BST (Implemented as set in C++ and TreeSet in Java). In a Self Balancing BST, we can do both insert and closest greater operations in O(Log n) time.

We use lower_bound() in C++ to find closest greater element. This function works in Log n time for a set.

C++

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// C++ implementation of efficient algorithm to find
// greater or same element on left side
#include <iostream>
#include <set>
using namespace std;
  
// Prints greater elements on left side of every element
void printPrevGreater(int arr[], int n)
{
    set<int> s;
    for (int i = 0; i < n; i++) {
  
        // First search in set
        auto it = s.lower_bound(arr[i]);
        if (it == s.end())   // If no greater found
            cout << "-1" << " ";
        else                   
            cout << *it << " ";        
  
        // Then insert
        s.insert(arr[i]);
    }
}
  
/* Driver program to test insertion sort */
int main()
{
    int arr[] = {10, 5, 11, 10, 20, 12};
    int n = sizeof(arr) / sizeof(arr[0]);
    printPrevGreater(arr, n);
    return 0;
}

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Java

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// Java implementation of efficient algorithm 
// to find greater or same element on left side 
import java.util.HashSet;
  
class GFG 
{
  
    // To get the minimum vaue from a set
    static int minimum(HashSet<Integer> ss)
    {
        int min = Integer.MAX_VALUE;
  
        for (int i : ss)
            if (i < min)
                min = i;
  
        return min;
    }
  
    // Prints greater elements on left side
    // of every element
    static void printPrevGreater(int[] arr, int n) 
    {
        HashSet<Integer> s = new HashSet<>();
  
        for (int i = 0; i < n; i++)
        {
            HashSet<Integer> ss = new HashSet<>();
  
            // First search in set
            for (int x : s)
                if (x >= arr[i])
                    ss.add(x);
  
            if (ss.size() == 0) // If no greater found
                System.out.print(-1 + " ");
            else
                System.out.print(minimum(ss) + " ");
  
            // Then insert
            s.add(arr[i]);
        }
    }
  
    // Driver Code
    public static void main(String[] args) 
    {
        int[] arr = { 10, 5, 11, 10, 20, 12 };
        int n = arr.length;
        printPrevGreater(arr, n);
    }
}
  
// This code is contributed by
// sanjeev2552

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Python3

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# Python3 implementation of efficient algorithm 
# to find greater or same element on left side 
  
# Prints greater elements 
# on left side of every element 
def printPrevGreater(arr, n): 
  
    s = set()
    for i in range(0, n): 
  
        # First search in set 
        it = [x for x in s if x >= arr[i]] 
        if len(it) == 0: # If no greater found 
            print("-1", end = " "
        else:                 
            print(min(it), end = " ")         
  
        # Then insert 
        s.add(arr[i]) 
  
# Driver Code
if __name__ == "__main__"
  
    arr = [10, 5, 11, 10, 20, 12
    n = len(arr) 
    printPrevGreater(arr, n) 
      
# This code is contributed by Rituraj Jain

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Output:

-1 10 -1 10 -1 20

Time Complexity : O(n Log n)



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Improved By : rituraj_jain, sanjeev2552