Given an array, find an element before which all elements are smaller than it, and after which all are greater than it. Return the index of the element if there is such an element, otherwise, return -1.
Examples:
Input: arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9};
Output: 4
Explanation: All elements on left of arr[4] are smaller than it
and all elements on right are greater.Input: arr[] = {5, 1, 4, 4};
Output: -1
Explanation : No such index exits.
Expected time complexity: O(n).
A simple solution is to consider every element one by one. For every element, compare it with all elements on the left and all elements on right. The time complexity of this solution is O(n2).
Code-
C++
// C++ program to find the element which is greater than// all left elements and smaller than all right elements.#include <bits/stdc++.h>using namespace std;//Function to checkbool check(int arr[], int n,int ind){ int i=ind-1; int j=ind+1; while(i>=0){ if(arr[i]>arr[ind]){return false;} i--; } while(j<n){ if(arr[j]<arr[ind]){return false;} j++; } return true;}// Function to return the index of the element which is greater than// all left elements and smaller than all right elements.int findElement(int arr[], int n){ // Traverse array from 1st to n-1 th index because //Extrem elements can't be aur answer for (int i=1; i<n-1; i++) { if(check(arr,n,i)){return i;} } // If there was no element matching criteria return -1;}// Driver programint main(){ int arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9}; int n = sizeof arr / sizeof arr[0]; cout << "Index of the element is " << findElement(arr, n); return 0;} |
Java
// Java program to find the element which is greater than// all left elements and smaller than all right elements.import java.util.*;class Main { // Function to check static boolean check(int arr[], int n, int ind) { int i = ind - 1; int j = ind + 1; while (i >= 0) { if (arr[i] > arr[ind]) { return false; } i--; } while (j < n) { if (arr[j] < arr[ind]) { return false; } j++; } return true; } // Function to return the index of the element which is // greater than all left elements and smaller than all // right elements. static int findElement(int arr[], int n) { // Traverse array from 1st to n-1 th index because // Extrem elements can't be aur answer for (int i = 1; i < n - 1; i++) { if (check(arr, n, i)) { return i; } } // If there was no element matching criteria return -1; } // Driver program public static void main(String[] args) { int arr[] = { 5, 1, 4, 3, 6, 8, 10, 7, 9 }; int n = arr.length; System.out.println("Index of the element is " + findElement(arr, n)); }} |
Python3
# python3 program to find the element which is greater than# all left elements and smaller than all right elements.# Function to checkdef check(arr, n, ind): i = ind - 1 j = ind + 1 while i >= 0: if arr[i] > arr[ind]: return False i -= 1 while j < n: if arr[j] < arr[ind]: return False j += 1 return True# Function to return the index of the element which is greater than# all left elements and smaller than all right elements.def findElement(arr, n): # Traverse array from 1st to n-1 th index because # Extrem elements can't be aur answer for i in range(1, n - 1): if check(arr, n, i): return i # If there was no element matching criteria return -1arr = [5, 1, 4, 3, 6, 8, 10, 7, 9]n = len(arr)print("Index of the element is", findElement(arr, n)) |
C#
using System;public class MainClass { public static bool Check(int[] arr, int n, int ind) { int i = ind - 1; int j = ind + 1; while (i >= 0) { if (arr[i] > arr[ind]) { return false; } i--; } while (j < n) { if (arr[j] < arr[ind]) { return false; } j++; } return true; } public static int FindElement(int[] arr, int n) { // Traverse array from 1st to n-1 th index because // Extrem elements can't be aur answer for (int i = 1; i < n - 1; i++) { if (Check(arr, n, i)) { return i; } } // If there was no element matching criteria return -1; } public static void Main() { int[] arr = { 5, 1, 4, 3, 6, 8, 10, 7, 9 }; int n = arr.Length; Console.WriteLine("Index of the element is " + FindElement(arr, n)); }} |
Javascript
// Function to checkfunction check(arr, n, ind) { let i = ind - 1; let j = ind + 1; while (i >= 0) { if (arr[i] > arr[ind]) { return false; } i--; } while (j < n) { if (arr[j] < arr[ind]) { return false; } j++; } return true;}// Function to return the index of the element which is greater than// all left elements and smaller than all right elements.function findElement(arr, n) { // Traverse array from 1st to n-1 th index because // Extrem elements can't be our answer for (let i = 1; i < n - 1; i++) { if (check(arr, n, i)) { return i; } } // If there was no element matching criteria return -1;}// Driver programlet arr = [5, 1, 4, 3, 6, 8, 10, 7, 9];let n = arr.length;console.log("Index of the element is " + findElement(arr, n));// This code is contributed by sarojmcy2e |
Output
Index of the element is 4
Time Complexity: O(n2), Time complexity of the given program is O(n^2) as there are two nested while loops in the check function, which are iterating over at most n-2 elements each, and they are being called for each element in the array except the first and last elements.
Auxiliary Space: O(1), Space complexity of the program is O(1) as no extra space is being used, except for the input array and some integer variables used for indexing and loop control.
An Efficient Solution can solve this problem in O(n) time using O(n) extra space. Below is the detailed solution.
- Create two arrays leftMax[] and rightMin[].
- Traverse input array from left to right and fill leftMax[] such that leftMax[i] contains a maximum element from 0 to i-1 in the input array.
- Traverse input array from right to left and fill rightMin[] such that rightMin[i] contains a minimum element from to n-1 to i+1 in the input array.
- Traverse input array. For every element arr[i], check if arr[i] is greater than leftMax[i] and smaller than rightMin[i]. If yes, return i.
Further Optimization to the above approach is to use only one extra array and traverse input array only twice. The first traversal is the same as above and fills leftMax[]. Next traversal traverses from the right and keeps track of the minimum. The second traversal also finds the required element.
Below image is a dry run of the above approach:
Below is the implementation of the above approach.
C++
// C++ program to find the element which is greater than// all left elements and smaller than all right elements.#include <bits/stdc++.h>using namespace std;// Function to return the index of the element which is greater than// all left elements and smaller than all right elements.int findElement(int arr[], int n){ // leftMax[i] stores maximum of arr[0..i-1] int leftMax[n]; leftMax[0] = INT_MIN; // Fill leftMax[1..n-1] for (int i = 1; i < n; i++) leftMax[i] = max(leftMax[i-1], arr[i-1]); // Initialize minimum from right int rightMin = INT_MAX; // Traverse array from right for (int i=n-1; i>=0; i--) { // Check if we found a required element if (leftMax[i] < arr[i] && rightMin > arr[i]) return i; // Update right minimum rightMin = min(rightMin, arr[i]); } // If there was no element matching criteria return -1;}// Driver programint main(){ int arr[] = {5, 1, 4, 3, 6, 8, 10, 7, 9}; int n = sizeof arr / sizeof arr[0]; cout << "Index of the element is " << findElement(arr, n); return 0;} |
Java
// Java program to find the element which is greater than// all left elements and smaller than all right elements.import java.io.*;import java.util.*;public class GFG { static int findElement(int[] arr, int n) { // leftMax[i] stores maximum of arr[0..i-1] int[] leftMax = new int[n]; leftMax[0] = Integer.MIN_VALUE; // Fill leftMax[1..n-1] for (int i = 1; i < n; i++) leftMax[i] = Math.max(leftMax[i - 1], arr[i - 1]); // Initialize minimum from right int rightMin = Integer.MAX_VALUE; // Traverse array from right for (int i = n - 1; i >= 0; i--) { // Check if we found a required element if (leftMax[i] < arr[i] && rightMin > arr[i]) return i; // Update right minimum rightMin = Math.min(rightMin, arr[i]); } // If there was no element matching criteria return -1; } // Driver code public static void main(String args[]) { int[] arr = {5, 1, 4, 3, 6, 8, 10, 7, 9}; int n = arr.length; System.out.println("Index of the element is " + findElement(arr, n)); } // This code is contributed // by rachana soma} |
Python3
# Python3 program to find the element which is greater than# all left elements and smaller than all right elements.def findElement(arr, n): # leftMax[i] stores maximum of arr[0..i-1] leftMax = [None] * n leftMax[0] = arr[0] # Fill leftMax[1..n-1] for i in range(1, n): leftMax[i] = max(leftMax[i-1], arr[i-1]) # Initialize minimum from right rightMin = [None]*n rightMin[n-1] = arr[n-1] # Fill rightMin for i in range(n-2, -1, -1): rightMin[i] = min(rightMin[i+1], arr[i]) # Traverse array from right for i in range(1, n-1): # Check if we found a required element # for ith element, it should be more than maximum of array # elements [0....i-1] and should be less than the minimum of # [i+1.....n-1] array elements if leftMax[i-1] <= arr[i] and arr[i] <= rightMin[i+1]: return i # If there was no element matching criteria return -1# Driver programif __name__ == "__main__": arr = [5, 1, 4, 3, 6, 8, 10, 7, 9] n = len(arr) print("Index of the element is", findElement(arr, n))# This code is contributed by Rituraj Jain |
C#
// C# program to find the element which is greater than// all left elements and smaller than all right elements.using System;class GFG{static int findElement(int[] arr, int n){ // leftMax[i] stores maximum of arr[0..i-1] int[] leftMax = new int[n]; leftMax[0] = int.MinValue; // Fill leftMax[1..n-1] for (int i = 1; i < n; i++) leftMax[i] = Math.Max(leftMax[i - 1], arr[i - 1]); // Initialize minimum from right int rightMin = int.MaxValue; // Traverse array from right for (int i=n-1; i>=0; i--) { // Check if we found a required element if (leftMax[i] < arr[i] && rightMin > arr[i]) return i; // Update right minimum rightMin = Math.Min(rightMin, arr[i]); } // If there was no element matching criteria return -1;}// Driver programpublic static void Main(){ int[] arr = {5, 1, 4, 3, 6, 8, 10, 7, 9}; int n = arr.Length; Console.Write( "Index of the element is " + findElement(arr, n));}}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP program to find the element // which is greater than all left // elements and smaller than all // right elements. function findElement($arr, $n) { // leftMax[i] stores maximum // of arr[0..i-1] $leftMax = array(0); $leftMax[0] = PHP_INT_MIN; // Fill leftMax[1..n-1] for ($i = 1; $i < $n; $i++) $leftMax[$i] = max($leftMax[$i - 1], $arr[$i - 1]); // Initialize minimum from right $rightMin = PHP_INT_MAX; // Traverse array from right for ($i = $n - 1; $i >= 0; $i--) { // Check if we found a required // element if ($leftMax[$i] < $arr[$i] && $rightMin > $arr[$i]) return $i; // Update right minimum $rightMin = min($rightMin, $arr[$i]); } // If there was no element // matching criteria return -1; } // Driver Code $arr = array(5, 1, 4, 3, 6, 8, 10, 7, 9); $n = count($arr);echo "Index of the element is " , findElement($arr, $n); // This code is contributed// by Sach_Code?> |
Javascript
<script>// Javascript program to find the element// which is greater than all left elements// and smaller than all right elements.// Function to return the index of the // element which is greater than all // left elements and smaller than all right elements.function findElement(arr, n) { // leftMax[i] stores maximum of arr[0..i-1] var leftMax = Array(n).fill(0); leftMax[0] = Number.MIN_VALUE; // Fill leftMax[1..n-1] for(i = 1; i < n; i++) leftMax[i] = Math.max(leftMax[i - 1], arr[i - 1]); // Initialize minimum from right var rightMin = Number.MAX_VALUE; // Traverse array from right for(i = n - 1; i >= 0; i--) { // Check if we found a required element if (leftMax[i] < arr[i] && rightMin > arr[i]) return i; // Update right minimum rightMin = Math.min(rightMin, arr[i]); } // If there was no element // matching criteria return -1;}// Driver codevar arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ];var n = arr.length;document.write("Index of the element is " + findElement(arr, n));// This code is contributed by aashish1995 </script> |
Output:
Index of the element is 4
Time Complexity: O(n), The program uses two loops to traverse the input array, one from left to right and another from right to left. The time complexity of the first loop is O(n) and that of the second loop is also O(n), so the overall time complexity of the program is O(n).
Auxiliary Space: O(n), The program uses an extra array of size n to store the maximum of all left elements, so the space complexity of the program is O(n).
Thanks to Gaurav Ahirwar for suggesting the above solution.
Space Optimized Approach:
C++
// C++ program to find the element which is greater than// all left elements and smaller than all right elements.#include <bits/stdc++.h>using namespace std;int findElement(int a[], int n){ // Base case if (n == 1 || n == 2) { return -1; } // 1.element is the possible candidate for the solution // of the problem. // 2.idx is the index of the possible // candidate. // 3.maxx is the value which is maximum on the // left side of the array. // 4.bit tell whether the loop is // terminated from the if condition or from the else // condition when loop do not satisfied the condition. // 5.check is the variable which tell whether the // element is updated or not int element = a[0], maxx = a[0], bit = -1, check = 0; int idx = -1; // The extreme two of the array can not be the solution // Therefore iterate the loop from i = 1 to < n-1 for (int i = 1; i < (n - 1);) { // here we find the possible candidate where Element // with left side smaller and right side greater. // when the if condition fail we check and update in // else condition. if (a[i] < maxx && i < (n - 1)) { i++; bit = 0; } // here we update the possible element if the // element is greater than the maxx (maximum element // so far). In while loop we sur-pass the value which // is greater than the element else { if (a[i] >= maxx) { element = a[i]; idx = i; check = 1; maxx = a[i]; } if (check == 1) { i++; } bit = 1; while (a[i] >= element && i < (n - 1)) { if (a[i] > maxx) { maxx = a[i]; } i++; } check = 0; } } // checking for the last value and whether the loop is // terminated from else or if block. if (element <= a[n - 1] && bit == 1) { return idx; } else { return -1; }}// Driver Codeint main(){ int arr[] = { 5, 1, 4, 3, 6, 8, 10, 7, 9 }; int n = sizeof arr / sizeof arr[0]; // Function Call cout << "Index of the element is " << findElement(arr, n); return 0;} |
Java
// Java program to find the element // which is greater than all left// elements and smaller than all// right elements.class GFG{ static int findElement(int []a, int n){ // Base case if (n == 1 || n == 2) { return -1; } // 1.element is the possible candidate for // the solution of the problem. // 2.idx is the index of the possible // candidate. // 3.maxx is the value which is maximum on the // left side of the array. // 4.bit tell whether the loop is // terminated from the if condition or from // the else condition when loop do not // satisfied the condition. // 5.check is the variable which tell whether the // element is updated or not int element = a[0], maxx = a[0], bit = -1, check = 0; int idx = -1; // The extreme two of the array can // not be the solution. Therefore // iterate the loop from i = 1 to < n-1 for(int i = 1; i < (n - 1);) { // Here we find the possible candidate // where Element with left side smaller // and right side greater. When the if // condition fail we check and update in // else condition. if (a[i] < maxx && i < (n - 1)) { i++; bit = 0; } // Here we update the possible element // if the element is greater than the // maxx (maximum element so far). In // while loop we sur-pass the value which // is greater than the element else { if (a[i] >= maxx) { element = a[i]; idx = i; check = 1; maxx = a[i]; } if (check == 1) { i++; } bit = 1; while (a[i] >= element && i < (n - 1)) { if (a[i] > maxx) { maxx = a[i]; } i++; } check = 0; } } // Checking for the last value and whether // the loop is terminated from else or // if block. if (element <= a[n - 1] && bit == 1) { return idx; } else { return -1; } }// Driver codepublic static void main(String []args){ int []arr = { 5, 1, 4, 3, 6, 8, 10, 7, 9 }; int n = arr.length; System.out.println("Index of the element is " + findElement(arr, n));}}// This code is contributed by avanitrachhadiya2155 |
Python3
# Python3 program to find the element which # is greater than all left elements and # smaller than all right elements.def findElement (a, n): # Base case if (n == 1 or n == 2): return -1 # 1. element is the possible candidate # for the solution of the problem # 2. idx is the index of the # possible candidate # 3. maxx is the value which is maximum # on the left side of the array # 4. bit tell whether the loop is # terminated from the if condition or # from the else condition when loop do # not satisfied the condition. # 5. check is the variable which tell # whether the element is updated or not element, maxx, bit = a[0], a[0], -1 check = 0 idx = -1 # The extreme of the array can't be # the solution Therefore iterate # the loop from i = 1 to < n-1 i = 1 while (i < (n - 1)): # Here we find the possible candidate # where element with left side smaller # and right side greater. when the if # condition fail we check and update # in else condition if (a[i] < maxx and i < (n - 1)): i += 1 bit = 0 # Here we update the possible element # if the element is greater than the # maxx (maximum element so far). In # while loop we sur-pass the value # which is greater than the element else: if (a[i] >= maxx): element = a[i] idx = i check = 1 maxx = a[i] if (check == 1): i += 1 bit = 1 while (a[i] >= element and i < (n - 1)): if (a[i] > maxx): maxx = a[i] i += 1 check = 0 # Checking for the last value and whether # the loop is terminated from else or # if block if (element <= a[n - 1] and bit == 1): return idx else: return -1# Driver Codeif __name__ == '__main__': arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ] n = len(arr) # Function call print("Index of the element is", findElement(arr, n))# This code is contributed by himanshu77 |
C#
// C# program to find the element // which is greater than all left// elements and smaller than all// right elements.using System; class GFG{ static int findElement(int []a, int n){ // Base case if (n == 1 || n == 2) { return -1; } // 1.element is the possible candidate for // the solution of the problem. // 2.idx is the index of the possible // candidate. // 3.maxx is the value which is maximum on the // left side of the array. // 4.bit tell whether the loop is // terminated from the if condition or from // the else condition when loop do not // satisfied the condition. // 5.check is the variable which tell whether the // element is updated or not int element = a[0], maxx = a[0], bit = -1, check = 0; int idx = -1; // The extreme two of the array can // not be the solution. Therefore // iterate the loop from i = 1 to < n-1 for(int i = 1; i < (n - 1);) { // Here we find the possible candidate // where Element with left side smaller // and right side greater. When the if // condition fail we check and update in // else condition. if (a[i] < maxx && i < (n - 1)) { i++; bit = 0; } // Here we update the possible element // if the element is greater than the // maxx (maximum element so far). In // while loop we sur-pass the value which // is greater than the element else { if (a[i] >= maxx) { element = a[i]; idx = i; check = 1; maxx = a[i]; } if (check == 1) { i++; } bit = 1; while (a[i] >= element && i < (n - 1)) { if (a[i] > maxx) { maxx = a[i]; } i++; } check = 0; } } // Checking for the last value and whether // the loop is terminated from else or // if block. if (element <= a[n - 1] && bit == 1) { return idx; } else { return -1; }}// Driver codepublic static void Main(string[] args){ int []arr = { 5, 1, 4, 3, 6, 8, 10, 7, 9 }; int n = arr.Length; // Function Call Console.Write("Index of the element is " + findElement(arr, n));}}// This code is contributed by rutvik_56 |
Javascript
<script>// javascript program to find the element // which is greater than all left// elements and smaller than all// right elements. function findElement(a , n){ // Base case if (n == 1 || n == 2) { return -1; } // 1.element is the possible candidate for // the solution of the problem. // 2.idx is the index of the possible // candidate. // 3.maxx is the value which is maximum on the // left side of the array. // 4.bit tell whether the loop is // terminated from the if condition or from // the else condition when loop do not // satisfied the condition. // 5.check is the variable which tell whether the // element is updated or not var element = a[0], maxx = a[0], bit = -1, check = 0; var idx = -1; // The extreme two of the array can // not be the solution. Therefore // iterate the loop from i = 1 to < n-1 for (i = 1; i < (n - 1);) { // Here we find the possible candidate // where Element with left side smaller // and right side greater. When the if // condition fail we check and update in // else condition. if (a[i] < maxx && i < (n - 1)) { i++; bit = 0; } // Here we update the possible element // if the element is greater than the // maxx (maximum element so far). In // while loop we sur-pass the value which // is greater than the element else { if (a[i] >= maxx) { element = a[i]; idx = i; check = 1; maxx = a[i]; } if (check == 1) { i++; } bit = 1; while (a[i] >= element && i < (n - 1)) { if (a[i] > maxx) { maxx = a[i]; } i++; } check = 0; } } // Checking for the last value and whether // the loop is terminated from else or // if block. if (element <= a[n - 1] && bit == 1) { return idx; } else { return -1; } } // Driver code var arr = [ 5, 1, 4, 3, 6, 8, 10, 7, 9 ]; var n = arr.length; document.write("Index of the element is " + findElement(arr, n));// This code is contributed by gauravrajput1 </script> |
Output
Index of the element is 4
Time Complexity: O(n), The time complexity of this program is O(n) where n is the size of the input array. This is because the program iterates through the array only once to find the element that satisfies the given condition.
Auxiliary Space: O(1), The space complexity of this program is O(1) because it uses only a constant amount of extra space to store some variables like element, maxx, bit, check, and idx, which are not dependent on the input size. Therefore, the space used by the program does not increase with the size of the input array.
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