Search for an element in a Mountain Array
Last Updated :
02 Aug, 2022
Given a mountain array arr[] and an integer X, the task is to find the smallest index of X in the given array. If no such index is found, print -1.
Examples:
Input: arr = {1, 2, 3, 4, 5, 3, 1}, X = 3
Output: 2
Explanation:
The smallest index of X(= 3) in the array is 2.
Therefore, the required output is 2.
Input: arr[] = {0, 1, 2, 4, 2, 1}, X = 3
Output: -1
Explanation: Since 3 does not exist in the array, the required output is -1.
Naive Approach: The simplest approach is to traverse the array and check if the element in the current index is equal to X or not. If found to be true, then print that index.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: The optimal idea to solve this problem is to use Binary Search. Follow the steps below to solve this problem:
Below is the implementation of the above approach:
C++
// CPP program for the above approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the index of
// the peak element in the array
int findPeak(vector<int> arr)
{
// Stores left most index in which
// the peak element can be found
int left = 0;
// Stores right most index in which
// the peak element can be found
int right = arr.size() - 1;
while (left < right)
{
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If element at mid is less than
// element at (mid + 1)
if (arr[mid] < arr[mid + 1])
{
// Update left
left = mid + 1;
}
else
{
// Update right
right = mid;
}
}
return left;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an ascending order
int BS(int X, int left, int right,
vector<int> arr)
{
while (left <= right)
{
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If X found at mid
if (arr[mid] == X)
{
return mid;
}
// If X is greater than mid
else if (X > arr[mid])
{
// Update left
left = mid + 1;
}
else
{
// Update right
right = mid - 1;
}
}
return -1;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an descending order
int reverseBS(int X, int left, int right, vector<int> arr)
{
while (left <= right)
{
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If X found at mid
if (arr[mid] == X)
{
return mid;
}
else if (X > arr[mid])
{
// Update right
right = mid - 1;
}
else
{
// Update left
left = mid + 1;
}
}
return -1;
}
// Function to find the smallest index of X
void findInMA(int X, vector<int> mountainArr)
{
// Stores index of peak element in array
int peakIndex = findPeak(mountainArr);
// Stores index of X in the array
int res = -1;
// If X greater than or equal to first element
// of array and less than the peak element
if (X >= mountainArr[0] && X <= mountainArr[peakIndex])
{
// Update res
res = BS(X, 0, peakIndex, mountainArr);
}
// If element not found on
// left side of peak element
if (res == -1)
{
// Update res
res = reverseBS(X, peakIndex + 1,
mountainArr.size() - 1,
mountainArr);
}
// Print res
cout<<res<<endl;
}
// Driver Code
int main()
{
// Given X
int X = 3;
// Given array
vector<int> list{1, 2, 3, 4, 5, 3, 1};
// Function Call
findInMA(X, list);
}
// This code is contributed by bgangwar59.
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the index of
// the peak element in the array
public static int findPeak(
ArrayList<Integer> arr)
{
// Stores left most index in which
// the peak element can be found
int left = 0;
// Stores right most index in which
// the peak element can be found
int right = arr.size() - 1;
while (left < right) {
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If element at mid is less than
// element at (mid + 1)
if (arr.get(mid) < arr.get(mid + 1)) {
// Update left
left = mid + 1;
}
else {
// Update right
right = mid;
}
}
return left;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an ascending order
static int BS(int X, int left, int right,
ArrayList<Integer> arr)
{
while (left <= right) {
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If X found at mid
if (arr.get(mid) == X) {
return mid;
}
// If X is greater than mid
else if (X > arr.get(mid)) {
// Update left
left = mid + 1;
}
else {
// Update right
right = mid - 1;
}
}
return -1;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an descending order
static int reverseBS(int X, int left, int right,
ArrayList<Integer> arr)
{
while (left <= right) {
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If X found at mid
if (arr.get(mid) == X) {
return mid;
}
else if (X > arr.get(mid)) {
// Update right
right = mid - 1;
}
else {
// Update left
left = mid + 1;
}
}
return -1;
}
// Function to find the smallest index of X
static void findInMA(int X,
ArrayList<Integer> mountainArr)
{
// Stores index of peak element in array
int peakIndex = findPeak(mountainArr);
// Stores index of X in the array
int res = -1;
// If X greater than or equal to first element
// of array and less than the peak element
if (X >= mountainArr.get(0)
&& X <= mountainArr.get(peakIndex)) {
// Update res
res = BS(X, 0, peakIndex, mountainArr);
}
// If element not found on
// left side of peak element
if (res == -1) {
// Update res
res = reverseBS(X, peakIndex + 1,
mountainArr.size() - 1,
mountainArr);
}
// Print res
System.out.println(res);
}
// Driver Code
public static void main(String[] args)
{
// Given X
int X = 3;
// Given array
ArrayList<Integer> list = new ArrayList<>(
Arrays.asList(1, 2, 3, 4, 5, 3, 1));
// Function Call
findInMA(X, list);
}
}
# Python3 program for the above approach
# Function to find the index of
# the peak element in the array
def findPeak(arr):
# Stores left most index in which
# the peak element can be found
left = 0
# Stores right most index in which
# the peak element can be found
right = len(arr) - 1
while (left < right):
# Stores mid of left and right
mid = left + (right - left) // 2
# If element at mid is less than
# element at(mid + 1)
if (arr[mid] < arr[(mid + 1)]):
# Update left
left = mid + 1
else:
# Update right
right = mid
return left
# Function to perform binary search in an
# a subarray if elements of the subarray
# are in an ascending order
def BS(X, left, right, arr):
while (left <= right):
# Stores mid of left and right
mid = left + (right - left) // 2
# If X found at mid
if (arr[mid] == X):
return mid
# If X is greater than mid
elif (X > arr[mid]):
# Update left
left = mid + 1
else:
# Update right
right = mid - 1
return -1
# Function to perform binary search in an
# a subarray if elements of the subarray
# are in an descending order
def reverseBS(X, left, right, arr):
while (left <= right):
# Stores mid of left and right
mid = left + (right - left) // 2
# If X found at mid
if (arr[mid] == X):
return mid
elif (X > arr[mid]):
# Update right
right = mid - 1
else:
# Update left
left = mid + 1
return -1
# Function to find the smallest index of X
def findInMA(X, mountainArr):
# Stores index of peak element in array
peakIndex = findPeak(mountainArr)
# Stores index of X in the array
res = -1
# If X greater than or equal to first element
# of array and less than the peak element
if (X >= mountainArr[0] and
X <= mountainArr[peakIndex]):
# Update res
res = BS(X, 0, peakIndex, mountainArr)
# If element not found on
# left side of peak element
if (res == -1):
# Update res
res = reverseBS(X, peakIndex + 1,
mountainArr.size() - 1,
mountainArr)
# Print res
print(res)
# Driver Code
if __name__ == "__main__":
# Given X
X = 3
# Given array
arr = [ 1, 2, 3, 4, 5, 3, 1 ]
# Function Call
findInMA(X, arr)
# This code is contributed by chitranayal
// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{
// Function to find the index of
// the peak element in the array
public static int findPeak(List<int> arr)
{
// Stores left most index in which
// the peak element can be found
int left = 0;
// Stores right most index in which
// the peak element can be found
int right = arr.Count - 1;
while (left < right)
{
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If element at mid is less than
// element at (mid + 1)
if (arr[mid] < arr[(mid + 1)])
{
// Update left
left = mid + 1;
}
else
{
// Update right
right = mid;
}
}
return left;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an ascending order
static int BS(int X, int left, int right,
List<int> arr)
{
while (left <= right)
{
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If X found at mid
if (arr[(mid)] == X)
{
return mid;
}
// If X is greater than mid
else if (X > arr[mid])
{
// Update left
left = mid + 1;
}
else
{
// Update right
right = mid - 1;
}
}
return -1;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an descending order
static int reverseBS(int X, int left, int right,
List<int> arr)
{
while (left <= right)
{
// Stores mid of left and right
int mid = left + (right - left) / 2;
// If X found at mid
if (arr[mid] == X)
{
return mid;
}
else if (X > arr[mid])
{
// Update right
right = mid - 1;
}
else
{
// Update left
left = mid + 1;
}
}
return -1;
}
// Function to find the smallest index of X
static void findInMA(int X,
List<int> mountainArr)
{
// Stores index of peak element in array
int peakIndex = findPeak(mountainArr);
// Stores index of X in the array
int res = -1;
// If X greater than or equal to first element
// of array and less than the peak element
if (X >= mountainArr[0]
&& X <= mountainArr[peakIndex])
{
// Update res
res = BS(X, 0, peakIndex, mountainArr);
}
// If element not found on
// left side of peak element
if (res == -1)
{
// Update res
res = reverseBS(X, peakIndex + 1,
mountainArr.Count - 1,
mountainArr);
}
// Print res
Console.WriteLine(res);
}
// Driver Code
public static void Main( )
{
// Given X
int X = 3;
// Given array
List<int> list = new List<int>(){1, 2, 3, 4, 5, 3, 1};
// Function Call
findInMA(X, list);
}
}
// This code is contributed by code_hunt.
<script>
// Javascript program for the above approach
// Function to find the index of
// the peak element in the array
function findPeak(arr)
{
// Stores left most index in which
// the peak element can be found
var left = 0;
// Stores right most index in which
// the peak element can be found
var right = arr.length - 1;
while (left < right)
{
// Stores mid of left and right
var mid = left + parseInt((right - left) / 2);
// If element at mid is less than
// element at (mid + 1)
if (arr[mid] < arr[mid + 1])
{
// Update left
left = mid + 1;
}
else
{
// Update right
right = mid;
}
}
return left;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an ascending order
function BS(X, left, right, arr)
{
while (left <= right)
{
// Stores mid of left and right
var mid = left + parseInt((right - left) / 2);
// If X found at mid
if (arr[mid] == X)
{
return mid;
}
// If X is greater than mid
else if (X > arr[mid])
{
// Update left
left = mid + 1;
}
else
{
// Update right
right = mid - 1;
}
}
return -1;
}
// Function to perform binary search in an
// a subarray if elements of the subarray
// are in an descending order
function reverseBS(X, left, right, arr)
{
while (left <= right)
{
// Stores mid of left and right
var mid = left + parseInt((right - left) / 2);
// If X found at mid
if (arr[mid] == X)
{
return mid;
}
else if (X > arr[mid])
{
// Update right
right = mid - 1;
}
else
{
// Update left
left = mid + 1;
}
}
return -1;
}
// Function to find the smallest index of X
function findInMA(X, mountainArr)
{
// Stores index of peak element in array
var peakIndex = findPeak(mountainArr);
// Stores index of X in the array
var res = -1;
// If X greater than or equal to first element
// of array and less than the peak element
if (X >= mountainArr[0] && X <= mountainArr[peakIndex])
{
// Update res
res = BS(X, 0, peakIndex, mountainArr);
}
// If element not found on
// left side of peak element
if (res == -1)
{
// Update res
res = reverseBS(X, peakIndex + 1,
mountainArr.length - 1,
mountainArr);
}
// Print res
document.write( res + "<br>");
}
// Driver Code
// Given X
var X = 3;
// Given array
var list = [1, 2, 3, 4, 5, 3, 1];
// Function Call
findInMA(X, list);
</script>
Time Complexity: O(Log(N))
Auxiliary Space: O(1)
Similar Reads
Binary Search Algorithm - Iterative and Recursive Implementation Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc
15 min read
What is Binary Search Algorithm? Binary Search is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half and the correct interval to find is decided based on the searched value and the mid value of the interval. Example of binary searchProperties of Binary Search:Binary search is performed o
1 min read
Time and Space Complexity Analysis of Binary Search Algorithm Time complexity of Binary Search is O(log n), where n is the number of elements in the array. It divides the array in half at each step. Space complexity is O(1) as it uses a constant amount of extra space. Example of Binary Search AlgorithmAspectComplexityTime ComplexityO(log n)Space ComplexityO(1)
3 min read
How to calculate "mid" or Middle Element Index in Binary Search? The most common method to calculate mid or middle element index in Binary Search Algorithm is to find the middle of the highest index and lowest index of the searchable space, using the formula mid = low + \frac{(high - low)}{2} Finding the middle index "mid" in Binary Search AlgorithmIs this method
6 min read
Variants of Binary Search
Variants of Binary SearchBinary search is very easy right? Well, binary search can become complex when element duplication occurs in the sorted list of values. It's not always the "contains or not" we search using Binary Search, but there are 5 variants such as below:1) Contains (True or False)Â 2) Index of first occurrence
15+ min read
Meta Binary Search | One-Sided Binary SearchMeta binary search (also called one-sided binary search by Steven Skiena in The Algorithm Design Manual on page 134) is a modified form of binary search that incrementally constructs the index of the target value in the array. Like normal binary search, meta binary search takes O(log n) time. Meta B
9 min read
The Ubiquitous Binary Search | Set 1We are aware of the binary search algorithm. Binary search is the easiest algorithm to get right. I present some interesting problems that I collected on binary search. There were some requests on binary search. I request you to honor the code, "I sincerely attempt to solve the problem and ensure th
15+ min read
Uniform Binary SearchUniform Binary Search is an optimization of Binary Search algorithm when many searches are made on same array or many arrays of same size. In normal binary search, we do arithmetic operations to find the mid points. Here we precompute mid points and fills them in lookup table. The array look-up gene
7 min read
Randomized Binary Search AlgorithmWe are given a sorted array A[] of n elements. We need to find if x is present in A or not.In binary search we always used middle element, here we will randomly pick one element in given range.In Binary Search we had middle = (start + end)/2 In Randomized binary search we do following Generate a ran
13 min read
Abstraction of Binary SearchWhat is the binary search algorithm? Binary Search Algorithm is used to find a certain value of x for which a certain defined function f(x) needs to be maximized or minimized. It is frequently used to search an element in a sorted sequence by repeatedly dividing the search interval into halves. Begi
7 min read
N-Base modified Binary Search algorithmN-Base modified Binary Search is an algorithm based on number bases that can be used to find an element in a sorted array arr[]. This algorithm is an extension of Bitwise binary search and has a similar running time. Examples: Input: arr[] = {1, 4, 5, 8, 11, 15, 21, 45, 70, 100}, target = 45Output:
10 min read
Implementation of Binary Search in different languages
C Program for Binary SearchIn this article, we will understand the binary search algorithm and how to implement binary search programs in C. We will see both iterative and recursive approaches and how binary search can reduce the time complexity of the search operation as compared to linear search.Table of ContentWhat is Bina
7 min read
C++ Program For Binary SearchBinary Search is a popular searching algorithm which is used for finding the position of any given element in a sorted array. It is a type of interval searching algorithm that keep dividing the number of elements to be search into half by considering only the part of the array where there is the pro
5 min read
C Program for Binary SearchIn this article, we will understand the binary search algorithm and how to implement binary search programs in C. We will see both iterative and recursive approaches and how binary search can reduce the time complexity of the search operation as compared to linear search.Table of ContentWhat is Bina
7 min read
Binary Search functions in C++ STL (binary_search, lower_bound and upper_bound)In C++, STL provide various functions like std::binary_search(), std::lower_bound(), and std::upper_bound() which uses the the binary search algorithm for different purposes. These function will only work on the sorted data.There are the 3 binary search function in C++ STL:Table of Contentbinary_sea
3 min read
Binary Search in JavaBinary search is a highly efficient searching algorithm used when the input is sorted. It works by repeatedly dividing the search range in half, reducing the number of comparisons needed compared to a linear search. Here, we are focusing on finding the middle element that acts as a reference frame t
6 min read
Binary Search (Recursive and Iterative) - PythonBinary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Below is the step-by-step algorithm for Binary Search:D
6 min read
Binary Search In JavaScriptBinary Search is a searching technique that works on the Divide and Conquer approach. It is used to search for any element in a sorted array. Compared with linear, binary search is much faster with a Time Complexity of O(logN), whereas linear search works in O(N) time complexityExamples: Input : arr
3 min read
Binary Search using pthreadBinary search is a popular method of searching in a sorted array or list. It simply divides the list into two halves and discards the half which has zero probability of having the key. On dividing, we check the midpoint for the key and use the lower half if the key is less than the midpoint and the
8 min read
Comparison with other Searching
Binary Search Intuition and Predicate Functions The binary search algorithm is used in many coding problems, and it is usually not very obvious at first sight. However, there is certainly an intuition and specific conditions that may hint at using binary search. In this article, we try to develop an intuition for binary search. Introduction to Bi
12 min read
Can Binary Search be applied in an Unsorted Array? Binary Search is a search algorithm that is specifically designed for searching in sorted data structures. This searching algorithm is much more efficient than Linear Search as they repeatedly target the center of the search structure and divide the search space in half. It has logarithmic time comp
9 min read
Find a String in given Array of Strings using Binary Search Given a sorted array of Strings arr and a string x, The task is to find the index of x in the array using the Binary Search algorithm. If x is not present, return -1.Examples:Input: arr[] = {"contribute", "geeks", "ide", "practice"}, x = "ide"Output: 2Explanation: The String x is present at index 2.
6 min read