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

Last Updated : 25 Feb, 2025

Given an array arr[] of size n. For each element in the array, find the value wise closest element to its left that is greater than or equal to the current element. If no such element exists, return -1 for that position.

Examples:

Input : arr[] = [10, 5, 11, 6, 20, 12]
Output : [-1, 10, -1, 10, -1, 20]
Explanation: The first element has nothing on the left side, so the answer for first is -1.
Second, element 5 has 10 on the left, so the answer is 10.
Third element 11 has nothing greater or the same, so the answer is -1.
Fourth element 6 has 10 as value wise closest, so the answer is 10.
Similarly, we get values for the fifth and sixth elements.
Input : arr[] = [1, 2, 3, 4, 5]
Output : [-1, -1, -1, -1, -1]
Explanation: The given array is arranged in strictly increasing order, and all the elements on left of any element are smaller. Thus no index has any value greater or equal in its left.
Input : arr[] = [5, 4, 3, 2, 1]
Output : [-1, 5, 4, 3, 2]

[Naive Approach] - Using Nested Loops - O(n^2) Time and O(1) Auxiliary Space

A idea is to run two nested loops. We pick an outer element one by one. For every picked element, we traverse toward the left of it and find the closest (value-wise) greater element.

C++

#include <bits/stdc++.h> using namespace std;  // Function to find the closest greater element // on the left of every element of the array vector<int> closestGreater(vector<int> &arr) {     int n = arr.size();      // to store the results     vector<int> res(n, -1);      for (int i = 1; i < n; i++) {          // stores the minimum difference         int diff = INT_MAX;          // traverse left side to i-th element         for (int j = 0; j < i; j++) {             if (arr[j] >= arr[i])                 diff = min(diff, arr[j] - arr[i]);         }          // if greater or equal value exists         if (diff != INT_MAX)             res[i] = arr[i] + diff;     }      return res; }  int main() {     vector<int> arr = { 10, 5, 11, 6, 20, 12 };     vector<int> res = closestGreater(arr);     for(auto i:res) {         cout<<i<<" ";     }     return 0; }

Java

// Function to find the closest greater element // on the left of every element of the array import java.util.*;  class GfG {      // Function to find the closest greater element     static int[] closestGreater(int[] arr) {         int n = arr.length;          // to store the results         int[] res = new int[n];         for (int i = 0; i < n; i++) {             res[i] = -1;         }          for (int i = 1; i < n; i++) {              // stores the minimum difference             int diff = Integer.MAX_VALUE;              // traverse left side to i-th element             for (int j = 0; j < i; j++) {                 if (arr[j] >= arr[i])                     diff = Math.min(diff, arr[j] - arr[i]);             }              // if greater or equal value exists             if (diff != Integer.MAX_VALUE)                 res[i] = arr[i] + diff;         }          return res;     }      public static void main(String[] args) {         int[] arr = {10, 5, 11, 6, 20, 12};         int[] res = closestGreater(arr);         for (int i : res) {             System.out.print(i + " ");         }     } }

Python

# Function to find the closest greater element # on the left of every element of the array def closestGreater(arr):     n = len(arr)      # to store the results     res = [-1] * n      for i in range(1, n):          # stores the minimum difference         diff = float('inf')          # traverse left side to i-th element         for j in range(i):             if arr[j] >= arr[i]:                 diff = min(diff, arr[j] - arr[i])          # if greater or equal value exists         if diff != float('inf'):             res[i] = arr[i] + diff      return res  if __name__ == "__main__":     arr = [10, 5, 11, 6, 20, 12]     res = closestGreater(arr)     for i in res:         print(i, end=" ")

// Function to find the closest greater element // on the left of every element of the array using System;  class GfG {      // Function to find the closest greater element     static int[] closestGreater(int[] arr) {         int n = arr.Length;          // to store the results         int[] res = new int[n];         for (int i = 0; i < n; i++) {             res[i] = -1;         }          for (int i = 1; i < n; i++) {              // stores the minimum difference             int diff = int.MaxValue;              // traverse left side to i-th element             for (int j = 0; j < i; j++) {                 if (arr[j] >= arr[i])                     diff = Math.Min(diff, arr[j] - arr[i]);             }              // if greater or equal value exists             if (diff != int.MaxValue)                 res[i] = arr[i] + diff;         }          return res;     }      static void Main() {         int[] arr = {10, 5, 11, 6, 20, 12};         int[] res = closestGreater(arr);         foreach (int i in res) {             Console.Write(i + " ");         }     } }

JavaScript

// Function to find the closest greater element // on the left of every element of the array function closestGreater(arr) {     let n = arr.length;      // to store the results     let res = new Array(n).fill(-1);      for (let i = 1; i < n; i++) {          // stores the minimum difference         let diff = Infinity;          // traverse left side to i-th element         for (let j = 0; j < i; j++) {             if (arr[j] >= arr[i])                 diff = Math.min(diff, arr[j] - arr[i]);         }          // if greater or equal value exists         if (diff !== Infinity)             res[i] = arr[i] + diff;     }      return res; }  let arr = [10, 5, 11, 6, 20, 12]; let res = closestGreater(arr); for (let i of res) {     process.stdout.write(i + " "); }

Output

-1 10 -1 10 -1 20

[Expected Approach] - Using Set - O(n Log n) Time and O(n) Auxiliary Space

The idea is to use Tree Set (or Sorted Set or Set in C++) to store the elements on left of every element, ensuring that elements are stored in sorted order. Thereafter, for each element of the array arr[], find the closest greater or equal value in O(Log n) Time.

C++

#include <bits/stdc++.h> using namespace std;  // Function to find the closest greater element // on the left of every element of the array vector<int> closestGreater(vector<int> &arr) {     int n = arr.size();      // to store the results     vector<int> res(n, -1);      // to store the elements in sorted order     set<int> s;      for (int i = 0; i < n; i++) {          // First search in set         auto it = s.lower_bound(arr[i]);          // If greater or equal value exists         if (it != s.end())              res[i] = *it;          // insert the current element in the set         s.insert(arr[i]);     }          return res; }  int main() {     vector<int> arr = { 10, 5, 11, 6, 20, 12 };     vector<int> res = closestGreater(arr);     for(auto i:res) {         cout<<i<<" ";     }     return 0; }

Java

// Function to find the closest greater element // on the left of every element of the array import java.util.TreeSet;  class GfG {      // Function to find the closest greater element     static int[] closestGreater(int[] arr) {         int n = arr.length;          // to store the results         int[] res = new int[n];         for (int i = 0; i < n; i++) {             res[i] = -1;         }          // to store the elements in sorted order         TreeSet<Integer> s = new TreeSet<>();          for (int i = 0; i < n; i++) {              // First search in set             Integer it = s.ceiling(arr[i]);              // If greater or equal value exists             if (it != null)                 res[i] = it;              // insert the current element in the set             s.add(arr[i]);         }                  return res;     }      public static void main(String[] args) {         int[] arr = {10, 5, 11, 6, 20, 12};         int[] res = closestGreater(arr);         for (int i : res) {             System.out.print(i + " ");         }     } }

Python

# Function to find the closest greater element # on the left of every element of the array import bisect  def closestGreater(arr):     n = len(arr)      # to store the results     res = [-1] * n      # to store the elements in sorted order     s = []      for i in range(n):          # First search in set         idx = bisect.bisect_left(s, arr[i])          # If greater or equal value exists         if idx < len(s):             res[i] = s[idx]          # insert the current element in the set         bisect.insort(s, arr[i])          return res  if __name__ == "__main__":     arr = [10, 5, 11, 6, 20, 12]     res = closestGreater(arr)     for i in res:         print(i, end=" ")

// Function to find the closest greater element // on the left of every element of the array using System; using System.Collections.Generic;  class GfG {      // Function to find the closest greater element     static int[] closestGreater(int[] arr) {         int n = arr.Length;          // to store the results         int[] res = new int[n];         for (int i = 0; i < n; i++) {             res[i] = -1;         }          // to store the elements in sorted order         SortedSet<int> s = new SortedSet<int>();          for (int i = 0; i < n; i++) {              // First search in set             int? it = null;             var view = s.GetViewBetween(arr[i], int.MaxValue);             if (view.Count > 0) {                 it = view.Min;             }              // If greater or equal value exists             if (it != null)                 res[i] = it.Value;              // insert the current element in the set             s.Add(arr[i]);         }                  return res;     }      static void Main() {         int[] arr = {10, 5, 11, 6, 20, 12};         int[] res = closestGreater(arr);         foreach (int i in res) {             Console.Write(i + " ");         }     } }

JavaScript

// Function to find the closest greater element // on the left of every element of the array function closestGreater(arr) {     let n = arr.length;      // to store the results     let res = new Array(n).fill(-1);      // to store the elements in sorted order     let s = [];      for (let i = 0; i < n; i++) {          // First search in set         let lo = 0, hi = s.length, candidate = null;         while (lo < hi) {             let mid = Math.floor((lo + hi) / 2);             if (s[mid] < arr[i])                 lo = mid + 1;             else {                 candidate = s[mid];                 hi = mid;             }         }          // If greater or equal value exists         if (candidate !== null)             res[i] = candidate;          // insert the current element in the set in sorted order         s.splice(lo, 0, arr[i]);     }      return res; }  let arr = [10, 5, 11, 6, 20, 12]; let res = closestGreater(arr); for (let i = 0; i < res.length; i++) {     process.stdout.write(res[i] + " "); }