Given an array of positive integers, the task is to find a point/element up to which elements form a strictly decreasing sequence first followed by a sequence of strictly increasing integers.
- Both of the sequences must at least be of length 2 (considering the common element).
- The last value of the decreasing sequence is the first value of the increasing sequence.
Note: Print “No such element exist”, if not found.
Examples:
Input: arr[] = {3, 2, 1, 2}
Output: 1
{3, 2, 1} is strictly decreasing
then {1, 2} is strictly increasing.
Input: arr[] = {3, 2, 1}
Output: No such element exist
Approach:
- First start traversing the array and keep traversing till the elements are in strictly decreasing order.
- If next elements is greater than the previous element store that element in point variable.
- Start traversing the elements from that point and keep traversing till the elements are in the strictly increasing order.
- After step 3, if all the elements get traversed then print that point.
- Else print “No such element exist.”
Note: If any of the two elements are equal then also print “No such element exist” as it should be strictly decreasing and increasing.
Below is the implementation of above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to check such sequenceint findElement(int a[], int n){ // set increasing/decreasing sequence counter to 1 int inc = 1, dec = 1, point; // for loop counter from 1 to last index of list for (int i = 1; i < n; i++) { // check if current int is // smaller than previous int if (a[i] < a[i - 1]) { // if inc is 1, i.e., increasing // seq. never started if (inc == 1) // increase dec by 1 dec++; else return -1; } // check if current int is greater than previous int else if (a[i] > a[i - 1]) { // if inc is 1, i.e., increasing seq. // starting for first time, if (inc == 1) // store a[i-1] in point point = a[i - 1]; // only if decreasing seq. minimum // count has been met, if (dec >= 2) // increase inc by 1 inc++; else // return -1 as decreasing seq. // min. count must be 2 return -1; } // check if current int is equal // to previous int, if so, else if (a[i] == a[i - 1]) return -1; } // check if inc >= 2 and dec >= 2 // return point if (inc >= 2 && dec >= 2) return point; // otherwise return -1 else return -1;}// Driver codeint main(){ int arr[] = { 3, 2, 1, 2 }; int n = sizeof(arr) / sizeof(arr[0]); int ele = findElement(arr, n); if (ele == -1) cout << "No such element exist"; else cout << ele; return 0;} |
Java
// Java implementation of above approach public class GFG { // Function to check such sequence static int findElement(int a[], int n) { // set increasing/decreasing sequence counter to 1 int inc = 1, dec = 1, point = 0; // for loop counter from 1 to last index of list for (int i = 1; i < n; i++) { // check if current int is // smaller than previous int if (a[i] < a[i - 1]) { // if inc is 1, i.e., increasing // seq. never started if (inc == 1) // increase dec by 1 dec++; else return -1; } // check if current int is greater than previous int else if (a[i] > a[i - 1]) { // if inc is 1, i.e., increasing seq. // starting for first time, if (inc == 1) // store a[i-1] in point point = a[i - 1]; // only if decreasing seq. minimum // count has been met, if (dec >= 2) // increase inc by 1 inc++; else // return -1 as decreasing seq. // min. count must be 2 return -1; } // check if current int is equal // to previous int, if so, else if (a[i] == a[i - 1]) return -1; } // check if inc >= 2 and dec >= 2 // return point if (inc >= 2 && dec >= 2) return point; // otherwise return -1 else return -1; } // Driver code public static void main(String args[]) { int arr[] = { 3, 2, 1, 2 }; int n = arr.length ; int ele = findElement(arr, n); if (ele == -1) System.out.println("No such element exist"); else System.out.println(ele); } // This Code is contributed by ANKITRAI1} |
Python
# Python implementation of above approachdef findElement(a, n): # set increasing sequence counter to 1 inc = 1 # set decreasing sequence counter to 1 dec = 1 # for loop counter from 1 to last # index of list [len(array)-1] for i in range(1, n): # check if current int is smaller than previous int if(a[i] < a[i-1]): # if inc is 1, i.e., increasing seq. never started, if inc == 1: # increase dec by 1 dec = dec + 1 else: # return -1 since if increasing seq. has started, # there cannot be a decreasing seq. in a valley return -1 # check if current int is greater than previous int elif(a[i] > a[i-1]): # if inc is 1, i.e., increasing seq. # starting for first time, if inc == 1: # store a[i-1] in point point = a[i-1] # only if decreasing seq. minimum count has been met, if dec >= 2: # increase inc by 1 inc = inc + 1 else: # return -1 as decreasing seq. min. count must be 2 return -1 # check if current int is equal to previous int, if so, elif(a[i] == a[i-1]): # return false as valley is always # strictly increasing / decreasing return -1 # check if inc >= 2 and dec >= 2, if(inc >= 2 and dec >= 2): # return point return point else: # otherwise return -1 return -1a = [2, 1, 2]n = len(a)ele = findElement(a, n)if ( ele == -1 ): print("No such element exist")else: print(ele) |
C#
// C# implementation of above approach using System;class GFG {// Function to check such sequence static int findElement(int []a, int n) { // set increasing/decreasing // sequence counter to 1 int inc = 1, dec = 1, point = 0; // for loop counter from 1 to // last index of list for (int i = 1; i < n; i++) { // check if current int is // smaller than previous int if (a[i] < a[i - 1]) { // if inc is 1, i.e., increasing // seq. never started if (inc == 1) // increase dec by 1 dec++; else return -1; } // check if current int is greater // than previous int else if (a[i] > a[i - 1]) { // if inc is 1, i.e., increasing // seq. starting for first time, if (inc == 1) // store a[i-1] in point point = a[i - 1]; // only if decreasing seq. // minimum count has been met, if (dec >= 2) // increase inc by 1 inc++; else // return -1 as decreasing // seq. min. count must be 2 return -1; } // check if current int is equal // to previous int, if so, else if (a[i] == a[i - 1]) return -1; } // check if inc >= 2 and // dec >= 2, return point if (inc >= 2 && dec >= 2) return point; // otherwise return -1 else return -1; } // Driver codepublic static void Main(){ int []arr = { 3, 2, 1, 2 }; int n = arr.Length ; int ele = findElement(arr, n); if (ele == -1) Console.WriteLine("No such element exist"); else Console.WriteLine(ele); }}// This code is contributed by Shashank |
PHP
<?php // PHP implementation of above approach// Function to check such sequencefunction findElement(&$a, $n){ // set increasing/decreasing // sequence counter to 1 $inc = 1; $dec = 1; // for loop counter from 1 // to last index of list for ($i = 1; $i < $n; $i++) { // check if current int is // smaller than previous int if ($a[$i] < $a[$i - 1]) { // if inc is 1, i.e., increasing // seq. never started if ($inc == 1) // increase dec by 1 $dec++; else return -1; } // check if current int is greater // than previous int else if ($a[$i] > $a[$i - 1]) { // if inc is 1, i.e., increasing // seq. starting for first time, if ($inc == 1) // store a[i-1] in point $point = $a[$i - 1]; // only if decreasing seq. // minimum count has been met, if ($dec >= 2) // increase inc by 1 $inc++; else // return -1 as decreasing // seq. min. count must be 2 return -1; } // check if current int is equal // to previous int, if so, else if ($a[$i] == $a[$i - 1]) return -1; } // check if inc >= 2 and // dec >= 2, return point if ($inc >= 2 && $dec >= 2) return $point; // otherwise return -1 else return -1;}// Driver code$arr = array(3, 2, 1, 2);$n = sizeof($arr);$ele = findElement($arr, $n);if ($ele == -1) echo "No such element exist";else echo $ele;// This code is contributed // by ChitraNayal?> |
Javascript
<script>// Java Script implementation of above approach // Function to check such sequence function findElement(a,n) { // set increasing/decreasing sequence counter to 1 let inc = 1, dec = 1, point = 0; // for loop counter from 1 to last index of list for (let i = 1; i < n; i++) { // check if current int is // smaller than previous int if (a[i] < a[i - 1]) { // if inc is 1, i.e., increasing // seq. never started if (inc == 1) // increase dec by 1 dec++; else return -1; } // check if current int is greater than previous int else if (a[i] > a[i - 1]) { // if inc is 1, i.e., increasing seq. // starting for first time, if (inc == 1) // store a[i-1] in point point = a[i - 1]; // only if decreasing seq. minimum // count has been met, if (dec >= 2) // increase inc by 1 inc++; else // return -1 as decreasing seq. // min. count must be 2 return -1; } // check if current int is equal // to previous int, if so, else if (a[i] == a[i - 1]) return -1; } // check if inc >= 2 and dec >= 2 // return point if (inc >= 2 && dec >= 2) return point; // otherwise return -1 else return -1; } // Driver code let arr = [3, 2, 1, 2 ]; let n = arr.length ; let ele = findElement(arr, n); if (ele == -1) document.write("No such element exist"); else document.write(ele); // This code is contributed by sravan kumar Gottumukkala</script> |
Output
1
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
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