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 the 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 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 closes, so the 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++
// 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; } |
Java
// Java implementation of efficient algorithm // to find greater or same element on left side import java.util.TreeSet; class GFG { // Prints greater elements on left side // of every element static void printPrevGreater(int[] arr, int n) { TreeSet<Integer> ts = new TreeSet<>(); for (int i = 0; i < n; i++) { Integer c = ts.ceiling(arr[i]); if (c == null) // If no greater found System.out.print(-1 + " "); else System.out.print(c + " "); // Then insert ts.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 |
Python3
# 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 |
C#
// C# implementation of efficient algorithm // to find greater or same element on left side using System; using System.Collections.Generic; class GFG { // To get the minimum value static int minimum(SortedSet<int> ss) { int min = int.MaxValue; foreach(int i in ss) if (i < min) min = i; return min; } // Prints greater elements on left side // of every element static void printPrevGreater(int[] arr, int n) { SortedSet<int> s = new SortedSet<int>(); for (int i = 0; i < n; i++) { SortedSet<int> ss = new SortedSet<int>(); // First search in set foreach(int x in s) if (x >= arr[i]) ss.Add(x); if (ss.Count == 0) // If no greater found Console.Write(-1 + " "); else Console.Write(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 PrinciRaj1992 |
-1 10 -1 10 -1 20
Time Complexity : O(n Log n)
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