Find first and last positions of an element in a sorted array
Last Updated : 04 Mar, 2025
Given a sorted array arr[] with possibly some duplicates, the task is to find the first and last occurrences of an element x in the given array.
Note: If the number x is not found in the array then return both the indices as -1.
Examples:
Input : arr[] = [1, 3, 5, 5, 5, 5, 67, 123, 125], x = 5
Output : 2 5
Explanation: First occurrence of 5 is at index 2 and last occurrence of 5 is at index 5
Input : arr[] = [1, 3, 5, 5, 5, 5, 7, 123, 125 ], x = 7
Output : 6 6
Explanation: First and last occurrence of 7 is at index 6
Input: arr[] = [1, 2, 3], x = 4
Output: -1 -1
Explanation: No occurrence of 4 in the array, so, output is [-1, -1]
[Naive Approach] – Using Iteration – O(n) Time and O(1) Space
The idea is to simply iterate on the elements of the given array and keep track of first and last occurrence of the value x.
C++ #include <bits/stdc++.h> using namespace std; // Function for finding first and last occurrence of x vector<int> find(vector<int> arr, int x) { int n = arr.size(); // Initialize first and last index int first = -1, last = -1; for (int i = 0; i < n; i++) { // If x is different, continue if (x != arr[i]) continue; // If first occurrence found if (first == -1) first = i; // Update last occurrence last = i; } vector<int> res = {first, last}; return res; } int main() { vector<int> arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; vector<int> res = find(arr, x); cout << res[0] << " " << res[1]; return 0; } // Function for finding first and last occurrence of x import java.util.*; class GfG { // Function for finding first and last occurrence of x static ArrayList<Integer> find(int[] arr, int x) { int n = arr.length; // Initialize first and last index int first = -1, last = -1; for (int i = 0; i < n; i++) { // If x is different, continue if (x != arr[i]) continue; // If first occurrence found if (first == -1) first = i; // Update last occurrence last = i; } ArrayList<Integer> res = new ArrayList<>(); res.add(first); res.add(last); return res; } public static void main(String[] args) { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; ArrayList<Integer> res = find(arr, x); System.out.println(res.get(0) + " " + res.get(1)); } } # Function for finding first and last occurrence of x def find(arr, x): n = len(arr) # Initialize first and last index first = -1 last = -1 for i in range(n): # If x is different, continue if x != arr[i]: continue # If first occurrence found if first == -1: first = i # Update last occurrence last = i res = [first, last] return res if __name__ == "__main__": arr = [1, 3, 5, 5, 5, 5, 67, 123, 125] x = 5 res = find(arr, x) print(res[0], res[1])
// Function for finding first and last occurrence of x using System; using System.Collections.Generic; class GfG { // Function for finding first and last occurrence of x static List<int> find(int[] arr, int x) { int n = arr.Length; // Initialize first and last index int first = -1, last = -1; for (int i = 0; i < n; i++) { // If x is different, continue if (x != arr[i]) continue; // If first occurrence found if (first == -1) first = i; // Update last occurrence last = i; } List<int> res = new List<int> { first, last }; return res; } static void Main() { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; List<int> res = find(arr, x); Console.WriteLine(res[0] + " " + res[1]); } } // Function for finding first and last occurrence of x function find(arr, x) { let n = arr.length; // Initialize first and last index let first = -1, last = -1; for (let i = 0; i < n; i++) { // If x is different, continue if (x !== arr[i]) continue; // If first occurrence found if (first === -1) first = i; // Update last occurrence last = i; } let res = [first, last]; return res; } let arr = [1, 3, 5, 5, 5, 5, 67, 123, 125]; let x = 5; let res = find(arr, x); console.log(res[0] + " " + res[1]); [Expected Approach] – Using Binary Search – O(log n) Time and O(1) Space
The idea is to find the first and last occurrence of a given number separately using binary search.
Follow the below given approach:
1. For the first occurrence of a number
- If (high >= low): Calculate mid = low + (high – low)/2;
- If ((mid == 0 || x > arr[mid-1]) && arr[mid] == x): return mid
- Else if (x > arr[mid]): return first(arr, (mid + 1), high, x, n);
- Else: return first(arr, low, (mid -1), x, n);
- Otherwise: return -1;
2. For the last occurrence of a number
- if (high >= low): calculate mid = low + (high – low)/2;
- if( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x ): return mid;
- else if(x < arr[mid]): return last(arr, low, (mid -1), x, n);
- else: return last(arr, (mid + 1), high, x, n);
- otherwise: return -1;
C++ #include <bits/stdc++.h> using namespace std; //Function for finding last occurrence of x int findLast(vector<int> arr, int x) { int n = arr.size(); // Initialize low and high index // to find the last occurrence int low = 0, high = n - 1; // Initialize last occurrence int last = -1; // Find last occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if (x == arr[mid]) { last = mid; low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if (x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return last; } // Function for finding first occurrence of x int findFirst(vector<int> arr, int x) { int n = arr.size(); // Initialize low and high index // to find the first occurrence int low = 0, high = n - 1; // Initialize first occurrence int first = -1; // Find first occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if (x == arr[mid]) { first = mid; high = mid - 1; } // If x is less than arr[mid], // then search in the left subarray else if (x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return first; } // Function for finding first and last occurrence of x vector<int> find(vector<int> arr, int x) { int n = arr.size(); // Find first and last index int first = findFirst(arr, x); int last = findLast(arr, x); vector<int> res = {first, last}; return res; } int main() { vector<int> arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; vector<int> res = find(arr, x); cout << res[0] << " " << res[1]; return 0; } // Function for finding first and last occurrence of x import java.util.*; class GfG { // Function for finding last occurrence of x static int findLast(int[] arr, int x) { int n = arr.length; // Initialize low and high index // to find the last occurrence int low = 0, high = n - 1; // Initialize last occurrence int last = -1; // Find last occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if(x == arr[mid]) { last = mid; low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if(x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return last; } // Function for finding first occurrence of x static int findFirst(int[] arr, int x) { int n = arr.length; // Initialize low and high index // to find the first occurrence int low = 0, high = n - 1; // Initialize first occurrence int first = -1; // Find first occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if(x == arr[mid]) { first = mid; high = mid - 1; } // If x is less than arr[mid], // then search in the left subarray else if(x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return first; } // Function for finding first and last occurrence of x static ArrayList<Integer> find(int[] arr, int x) { int n = arr.length; // Find first and last index int first = findFirst(arr, x); int last = findLast(arr, x); ArrayList<Integer> res = new ArrayList<>(); res.add(first); res.add(last); return res; } public static void main(String[] args) { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; ArrayList<Integer> res = find(arr, x); System.out.println(res.get(0) + " " + res.get(1)); } } # Function for finding first and last occurrence of x def findLast(arr, x): n = len(arr) # Initialize low and high index # to find the last occurrence low = 0 high = n - 1 # Initialize last occurrence last = -1 # Find last occurrence of x while low <= high: # Find the mid index mid = (low + high) // 2 # If x is equal to arr[mid] if x == arr[mid]: last = mid low = mid + 1 # If x is less than arr[mid], # then search in the left subarray elif x < arr[mid]: high = mid - 1 # If x is greater than arr[mid], # then search in the right subarray else: low = mid + 1 return last # Function for finding first occurrence of x def findFirst(arr, x): n = len(arr) # Initialize low and high index # to find the first occurrence low = 0 high = n - 1 # Initialize first occurrence first = -1 # Find first occurrence of x while low <= high: # Find the mid index mid = (low + high) // 2 # If x is equal to arr[mid] if x == arr[mid]: first = mid high = mid - 1 # If x is less than arr[mid], # then search in the left subarray elif x < arr[mid]: high = mid - 1 # If x is greater than arr[mid], # then search in the right subarray else: low = mid + 1 return first # Function for finding first and last occurrence of x def find(arr, x): n = len(arr) # Find first and last index first = findFirst(arr, x) last = findLast(arr, x) res = [first, last] return res if __name__ == "__main__": arr = [1, 3, 5, 5, 5, 5, 67, 123, 125] x = 5 res = find(arr, x) print(res[0], res[1])
// Function for finding first and last occurrence of x using System; using System.Collections.Generic; class GfG { // Function for finding last occurrence of x static int findLast(int[] arr, int x) { int n = arr.Length; // Initialize low and high index // to find the last occurrence int low = 0, high = n - 1; // Initialize last occurrence int last = -1; // Find last occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if(x == arr[mid]) { last = mid; low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if(x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return last; } // Function for finding first occurrence of x static int findFirst(int[] arr, int x) { int n = arr.Length; // Initialize low and high index // to find the first occurrence int low = 0, high = n - 1; // Initialize first occurrence int first = -1; // Find first occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if(x == arr[mid]) { first = mid; high = mid - 1; } // If x is less than arr[mid], // then search in the left subarray else if(x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return first; } // Function for finding first and last occurrence of x static List<int> find(int[] arr, int x) { int n = arr.Length; // Find first and last index int first = findFirst(arr, x); int last = findLast(arr, x); List<int> res = new List<int> { first, last }; return res; } static void Main() { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; List<int> res = find(arr, x); Console.WriteLine(res[0] + " " + res[1]); } } // Function for finding first and last occurrence of x function findFirst(arr, x) { let n = arr.length; // Initialize low and high index let low = 0, high = n - 1; // Initialize first occurrence let first = -1; // Find first occurrence of x while (low <= high) { // Find the mid index let mid = Math.floor((low + high) / 2); // If x is equal to arr[mid] if (x === arr[mid]) { first = mid; high = mid - 1; } // If x is less than arr[mid], // then search in the left subarray else if (x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return first; } function findLast(arr, x) { let n = arr.length; // Initialize low and high index let low = 0, high = n - 1; // Initialize last occurrence let last = -1; // Find last occurrence of x while (low <= high) { // Find the mid index let mid = Math.floor((low + high) / 2); // If x is equal to arr[mid] if (x === arr[mid]) { last = mid; low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if (x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return last; } function find(arr, x) { let n = arr.length; // Find first and last index let first = findFirst(arr, x); let last = findLast(arr, x); let res = [first, last]; return res; } let arr = [1, 3, 5, 5, 5, 5, 67, 123, 125]; let x = 5; let res = find(arr, x); console.log(res[0] + " " + res[1]); [Alternate Approach – 1] – Using Binary Search – O(log n) Time and O(1) Space
In the above given approach, we are creating two functions to find the first and last occurrence of the number separately. But instead of doing so, we can use a Boolean findFirst, which will be “true”, if we are searching for the first index and false otherwise. If arr[mid] == x, and findFirst is “true” set high = mid – 1, else set low = mid + 1. Everything else will work similar to above approach.
C++ #include <bits/stdc++.h> using namespace std; int search(vector<int> &arr, int x, bool findStart) { int n = arr.size(); // Initialize low and high index int low = 0, high = n - 1; // Initialize the index int ind = -1; // Find occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if (x == arr[mid]) { ind = mid; if(findStart == true) high = mid - 1; else low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if (x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return ind; } // Function for finding first and last occurrence of x vector<int> find(vector<int> &arr, int x) { // return index of first occurrence int first = search(arr, x, true); // return index of last occurrence int last = search(arr, x, false); vector<int> res = {first, last}; return res; } int main() { vector<int> arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; vector<int> res = find(arr, x); cout << res[0] << " " << res[1]; return 0; } // Function for finding first and last occurrence of x import java.util.*; class GfG { // Function for finding first and last occurrence of x static ArrayList<Integer> find(int[] arr, int x) { int n = arr.length; // return index of first occurrence int first = search(arr, x, true); // return index of last occurrence int last = search(arr, x, false); ArrayList<Integer> res = new ArrayList<>(); res.add(first); res.add(last); return res; } static int search(int[] arr, int x, boolean findStart) { int n = arr.length; // Initialize low and high index int low = 0, high = n - 1; // Initialize the index int ind = -1; // Find occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if(x == arr[mid]) { ind = mid; if(findStart) high = mid - 1; else low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if(x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return ind; } public static void main(String[] args) { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; ArrayList<Integer> res = find(arr, x); System.out.println(res.get(0) + " " + res.get(1)); } } # Function for finding first and last occurrence of x def search(arr, x, findStart): n = len(arr) # Initialize low and high index low = 0 high = n - 1 # Initialize the index ind = -1 # Find occurrence of x while low <= high: # Find the mid index mid = (low + high) // 2 # If x is equal to arr[mid] if x == arr[mid]: ind = mid if findStart: high = mid - 1 else: low = mid + 1 # If x is less than arr[mid], # then search in the left subarray elif x < arr[mid]: high = mid - 1 # If x is greater than arr[mid], # then search in the right subarray else: low = mid + 1 return ind # Function for finding first and last occurrence of x def find(arr, x): n = len(arr) # return index of first occurrence first = search(arr, x, True) # return index of last occurrence last = search(arr, x, False) res = [first, last] return res if __name__ == "__main__": arr = [1, 3, 5, 5, 5, 5, 67, 123, 125] x = 5 res = find(arr, x) print(res[0], res[1])
// Function for finding first and last occurrence of x using System; using System.Collections.Generic; class GfG { // Function for finding first and last occurrence of x static List<int> find(int[] arr, int x) { int n = arr.Length; // return index of first occurrence int first = search(arr, x, true); // return index of last occurrence int last = search(arr, x, false); List<int> res = new List<int> { first, last }; return res; } static int search(int[] arr, int x, bool findStart) { int n = arr.Length; // Initialize low and high index int low = 0, high = n - 1; // Initialize the index int ind = -1; // Find occurrence of x while(low <= high) { // Find the mid index int mid = (low + high) / 2; // If x is equal to arr[mid] if(x == arr[mid]) { ind = mid; if(findStart) high = mid - 1; else low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if(x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return ind; } static void Main() { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; List<int> res = find(arr, x); Console.WriteLine(res[0] + " " + res[1]); } } // Function for finding first and last occurrence of x function search(arr, x, findStart) { let n = arr.length; // Initialize low and high index let low = 0, high = n - 1; // Initialize the index let ind = -1; // Find occurrence of x while (low <= high) { // Find the mid index let mid = Math.floor((low + high) / 2); // If x is equal to arr[mid] if (x === arr[mid]) { ind = mid; if (findStart) high = mid - 1; else low = mid + 1; } // If x is less than arr[mid], // then search in the left subarray else if (x < arr[mid]) high = mid - 1; // If x is greater than arr[mid], // then search in the right subarray else low = mid + 1; } return ind; } function find(arr, x) { let n = arr.length; // return index of first occurrence let first = search(arr, x, true); // return index of last occurrence let last = search(arr, x, false); let res = [first, last]; return res; } let arr = [1, 3, 5, 5, 5, 5, 67, 123, 125]; let x = 5; let res = find(arr, x); console.log(res[0] + " " + res[1]); [Alternate Approach – 2] – Using Inbuilt Functions – O(log n) Time and O(1) Space
The idea is to use inbuilt functions to find the first and last occurrence of the number in the array arr[]. Like in C++ we can use lower and upper bound to find the last occurrence of the number.
C++ #include <bits/stdc++.h> using namespace std; // Function for finding first and last occurrence of x vector<int> find(vector<int> arr, int x) { int n = arr.size(); // return index of first number // greater than or equal to x int first = lower_bound(arr.begin(), arr.end(), x) - arr.begin(); // return index of first number // greater than x int last = upper_bound(arr.begin(), arr.end(), x) - arr.begin() - 1; // If x is not present if (first == n || arr[first] != x) { first = -1; last = -1; } vector<int> res = {first, last}; return res; } int main() { vector<int> arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; vector<int> res = find(arr, x); cout << res[0] << " " << res[1]; return 0; } // Function for finding first and last occurrence of x import java.util.*; class GfG { // Function for finding first and last occurrence of x static ArrayList<Integer> find(int[] arr, int x) { ArrayList<Integer> list = new ArrayList<>(); for (int val : arr) { list.add(val); } // return index of first number // greater than or equal to x int first = list.indexOf(x); // return index of first number // greater than x int last = list.lastIndexOf(x); // If x is not present if (first == -1) { last = -1; } ArrayList<Integer> res = new ArrayList<>(); res.add(first); res.add(last); return res; } public static void main(String[] args) { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; ArrayList<Integer> res = find(arr, x); System.out.println(res.get(0) + " " + res.get(1)); } } # Function for finding first and last occurrence of x import bisect def find(arr, x): n = len(arr) # return index of first number # greater than or equal to x first = bisect.bisect_left(arr, x) # return index of first number # greater than x last = bisect.bisect_right(arr, x) - 1 # If x is not present if first == n or arr[first] != x: first = -1 last = -1 res = [first, last] return res if __name__ == "__main__": arr = [1, 3, 5, 5, 5, 5, 67, 123, 125] x = 5 res = find(arr, x) print(res[0], res[1])
// Function for finding first and last occurrence of x using System; using System.Collections.Generic; class GfG { // Function for finding first and last occurrence of x static List<int> find(int[] arr, int x) { List<int> list = new List<int>(arr); // return index of first number // greater than or equal to x int first = list.IndexOf(x); // return index of first number // greater than x int last = list.LastIndexOf(x); // If x is not present if (first == -1) { last = -1; } List<int> res = new List<int> { first, last }; return res; } static void Main() { int[] arr = {1, 3, 5, 5, 5, 5, 67, 123, 125}; int x = 5; List<int> res = find(arr, x); Console.WriteLine(res[0] + " " + res[1]); } } // Function for finding first and last occurrence of x function find(arr, x) { let n = arr.length; // return index of first number // greater than or equal to x let first = arr.findIndex(e => e >= x); // return index of first number // greater than x let last = arr.lastIndexOf(x); // If x is not present if (first === -1 || arr[first] !== x) { first = -1; last = -1; } let res = [first, last]; return res; } let arr = [1, 3, 5, 5, 5, 5, 67, 123, 125]; let x = 5; let res = find(arr, x); console.log(res[0] + " " + res[1]); Extended Problem : Count number of occurrences in a sorted array
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