Given an array of sorted integers. We need to find the closest value to the given number. Array may contain duplicate values and negative numbers.
Examples:
Input : arr[] = {1, 2, 4, 5, 6, 6, 8, 9}
Target number = 11
Output : 9
9 is closest to 11 in given array
Input :arr[] = {2, 5, 6, 7, 8, 8, 9};
Target number = 4
Output : 5
5 is closest to 4 in given array
Input :arr[] = {2, 5, 6, 7, 8, 8, 9, 15, 19, 22, 32};
Target number = 34
Output : 32
32 is closest to 34 in given array
A simple solution is to traverse through the given array and keep track of absolute difference of current element with every element. Finally return the element that has minimum absolute difference.
An efficient solution is to use Binary Search.
C++
// CPP program to find element// closest to given target using binary search.#include <bits/stdc++.h>using namespace std;int getClosest(int, int, int);// Returns element closest to target in arr[]int findClosest(int arr[], int n, int target){ // Corner cases //left-side case if (target <= arr[0]) return arr[0]; //right-side case if (target >= arr[n - 1]) return arr[n - 1]; // Doing binary search int i = 0, j = n, mid = 0; while (i < j) { mid = (i + j) / 2; if (arr[mid] == target) return arr[mid]; /* If target is less than array element, then search in left */ if (target < arr[mid]) { // If target is greater than previous // to mid, return closest of two if (mid > 0 && target > arr[mid - 1]) return getClosest(arr[mid - 1], arr[mid], target); j = mid; } /* Repeat for left half */ // If target is greater than mid else { if (mid < n - 1 && target < arr[mid + 1]) return getClosest(arr[mid], arr[mid + 1], target); // update i i = mid + 1; } } // Only single element left after search return arr[mid];}// Method to compare which one is the more close.// We find the closest by taking the difference// between the target and both values. It assumes// that val2 is greater than val1 and target lies// between these two.int getClosest(int val1, int val2, int target){ if (target - val1 >= val2 - target) return val2; else return val1;}// Driver codeint main(){ int arr[] = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int target = 11; cout << (findClosest(arr, n, target));}// This code is contributed bu Smitha Dinesh Semwal |
Java
// Java program to find element closest to given target using binary search.import java.util.*;import java.lang.*;import java.io.*;class FindClosestNumber { // Returns element closest to target in arr[] public static int findClosest(int arr[], int target) { int n = arr.length; // Corner cases if (target <= arr[0]) return arr[0]; if (target >= arr[n - 1]) return arr[n - 1]; // Doing binary search int i = 0, j = n, mid = 0; while (i < j) { mid = (i + j) / 2; if (arr[mid] == target) return arr[mid]; /* If target is less than array element, then search in left */ if (target < arr[mid]) { // If target is greater than previous // to mid, return closest of two if (mid > 0 && target > arr[mid - 1]) return getClosest(arr[mid - 1], arr[mid], target); /* Repeat for left half */ j = mid; } // If target is greater than mid else { if (mid < n-1 && target < arr[mid + 1]) return getClosest(arr[mid], arr[mid + 1], target); i = mid + 1; // update i } } // Only single element left after search return arr[mid]; } // Method to compare which one is the more close // We find the closest by taking the difference // between the target and both values. It assumes // that val2 is greater than val1 and target lies // between these two. public static int getClosest(int val1, int val2, int target) { if (target - val1 >= val2 - target) return val2; else return val1; } // Driver code public static void main(String[] args) { int arr[] = { 1, 2, 4, 5, 6, 6, 8, 9 }; int target = 11; System.out.println(findClosest(arr, target)); }} |
Python3
# Python3 program to find element# closest to given target using binary search.# Returns element closest to target in arr[]def findClosest(arr, n, target): # Corner cases if (target <= arr[0]): return arr[0] if (target >= arr[n - 1]): return arr[n - 1] # Doing binary search i = 0; j = n; mid = 0 while (i < j): mid = (i + j) // 2 if (arr[mid] == target): return arr[mid] # If target is less than array # element, then search in left if (target < arr[mid]) : # If target is greater than previous # to mid, return closest of two if (mid > 0 and target > arr[mid - 1]): return getClosest(arr[mid - 1], arr[mid], target) # Repeat for left half j = mid # If target is greater than mid else : if (mid < n - 1 and target < arr[mid + 1]): return getClosest(arr[mid], arr[mid + 1], target) # update i i = mid + 1 # Only single element left after search return arr[mid]# Method to compare which one is the more close.# We find the closest by taking the difference# between the target and both values. It assumes# that val2 is greater than val1 and target lies# between these two.def getClosest(val1, val2, target): if (target - val1 >= val2 - target): return val2 else: return val1# Driver codearr = [1, 2, 4, 5, 6, 6, 8, 9] n = len(arr)target = 11print(findClosest(arr, n, target))# This code is contributed by Smitha Dinesh Semwal |
C#
// C# program to find element // closest to given target using binary search.using System;class GFG{ // Returns element closest // to target in arr[] public static int findClosest(int []arr, int target) { int n = arr.Length; // Corner cases if (target <= arr[0]) return arr[0]; if (target >= arr[n - 1]) return arr[n - 1]; // Doing binary search int i = 0, j = n, mid = 0; while (i < j) { mid = (i + j) / 2; if (arr[mid] == target) return arr[mid]; /* If target is less than array element, then search in left */ if (target < arr[mid]) { // If target is greater // than previous to mid, // return closest of two if (mid > 0 && target > arr[mid - 1]) return getClosest(arr[mid - 1], arr[mid], target); /* Repeat for left half */ j = mid; } // If target is // greater than mid else { if (mid < n-1 && target < arr[mid + 1]) return getClosest(arr[mid], arr[mid + 1], target); i = mid + 1; // update i } } // Only single element // left after search return arr[mid]; } // Method to compare which one // is the more close We find the // closest by taking the difference // between the target and both // values. It assumes that val2 is // greater than val1 and target // lies between these two. public static int getClosest(int val1, int val2, int target) { if (target - val1 >= val2 - target) return val2; else return val1; } // Driver code public static void Main() { int []arr = {1, 2, 4, 5, 6, 6, 8, 9}; int target = 11; Console.WriteLine(findClosest(arr, target)); }}// This code is contributed by anuj_67. |
Javascript
<script>// JavaScript program to find element // closest to given target using binary search.// Returns element closest to target in arr[]function findClosest(arr, target){ let n = arr.length; // Corner cases if (target <= arr[0]) return arr[0]; if (target >= arr[n - 1]) return arr[n - 1]; // Doing binary search let i = 0, j = n, mid = 0; while (i < j) { mid = (i + j) / 2; if (arr[mid] == target) return arr[mid]; // If target is less than array // element,then search in left if (target < arr[mid]) { // If target is greater than previous // to mid, return closest of two if (mid > 0 && target > arr[mid - 1]) return getClosest(arr[mid - 1], arr[mid], target); // Repeat for left half j = mid; } // If target is greater than mid else { if (mid < n - 1 && target < arr[mid + 1]) return getClosest(arr[mid], arr[mid + 1], target); i = mid + 1; // update i } } // Only single element left after search return arr[mid];}// Method to compare which one is the more close// We find the closest by taking the difference// between the target and both values. It assumes// that val2 is greater than val1 and target lies// between these two.function getClosest(val1, val2, target){ if (target - val1 >= val2 - target) return val2; else return val1; }// Driver Codelet arr = [ 1, 2, 4, 5, 6, 6, 8, 9 ];let target = 11;document.write(findClosest(arr, target));// This code is contributed by code_hunt</script> |
PHP
<?php// PHP program to find element closest // to given target using binary search. // Returns element closest to target in arr[] function findClosest($arr, $n, $target) { // Corner cases if ($target <= $arr[0]) return $arr[0]; if ($target >= $arr[$n - 1]) return $arr[$n - 1]; // Doing binary search $i = 0; $j = $n; $mid = 0; while ($i < $j) { $mid = ($i + $j) / 2; if ($arr[$mid] == $target) return $arr[$mid]; /* If target is less than array element, then search in left */ if ($target < $arr[$mid]) { // If target is greater than previous // to mid, return closest of two if ($mid > 0 && $target > $arr[$mid - 1]) return getClosest($arr[$mid - 1], $arr[$mid], $target); /* Repeat for left half */ $j = $mid; } // If target is greater than mid else { if ($mid < $n - 1 && $target < $arr[$mid + 1]) return getClosest($arr[$mid], $arr[$mid + 1], $target); // update i $i = $mid + 1; } } // Only single element left after search return $arr[$mid]; } // Method to compare which one is the more close. // We find the closest by taking the difference // between the target and both values. It assumes // that val2 is greater than val1 and target lies // between these two. function getClosest($val1, $val2, $target) { if ($target - $val1 >= $val2 - $target) return $val2; else return $val1; } // Driver code $arr = array( 1, 2, 4, 5, 6, 6, 8, 9 ); $n = sizeof($arr); $target = 11; echo (findClosest($arr, $n, $target)); // This code is contributed by Sachin.?> |
Output
9
Time Complexity: O(log n) (Due to Binary Search)
Auxiliary Space: O(log n) (implicit stack is created due to recursion)
Approach 2: Using Two Pointers:
Another approach to solve this problem is to use two pointers technique, where we maintain two pointers left and right, and move them towards each other based on their absolute difference with target.
Initialize left = 0 and right = n-1, where n is the size of the array.
Loop while left < right
a. If the absolute difference between arr[left] and target is less than or equal to the absolute difference between arr[right] and target, move left pointer one step to the right, i.e. left++
b. Else, move right pointer one step to the left, i.e. right–
Return arr[left], which will be the element closest to the target.
C++
#include <bits/stdc++.h>using namespace std;int findClosest(int arr[], int n, int target){ int left = 0, right = n - 1; while (left < right) { if (abs(arr[left] - target) <= abs(arr[right] - target)) { right--; } else { left++; } } return arr[left];}int main(){ int arr[] = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = sizeof(arr) / sizeof(arr[0]); int target = 11; cout << findClosest(arr, n, target); return 0;} |
Java
import java.util.*;public class Main { public static int findClosest(int[] arr, int n, int target) { int left = 0, right = n - 1; while (left < right) { if (Math.abs(arr[left] - target) <= Math.abs(arr[right] - target)) { right--; } else { left++; } } return arr[left]; } public static void main(String[] args) { int[] arr = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = arr.length; int target = 11; System.out.println(findClosest(arr, n, target)); }} |
Python3
import sysdef findClosest(arr, n, target): left, right = 0, n - 1 while left < right: if abs(arr[left] - target) <= abs(arr[right] - target): right -= 1 else: left += 1 return arr[left]if __name__ == "__main__": arr = [1, 2, 4, 5, 6, 6, 8, 8, 9] n = len(arr) target = 11 print(findClosest(arr, n, target)) |
C#
using System;public class Program { static int FindClosest(int[] arr, int n, int target) { int left = 0, right = n - 1; while (left < right) { if (Math.Abs(arr[left] - target) <= Math.Abs(arr[right] - target)) { right--; } else { left++; } } return arr[left]; } static void Main(string[] args) { int[] arr = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = arr.Length; int target = 11; Console.WriteLine(FindClosest(arr, n, target)); }} |
Javascript
function findClosest(arr, n, target) { let left = 0, right = n-1; while (left < right) { if (Math.abs(arr[left] - target) <= Math.abs(arr[right] - target)) { right--; } else { left++; } } return arr[left];}const arr = [1, 2, 4, 5, 6, 6, 8, 8, 9];const n = arr.length;const target = 11;console.log(findClosest(arr, n, target));// This code is contributed by shivhack999 |
Output
9
Time Complexity: O(N)
Auxiliary Space: O(1)
Approach 3: Recursive Approach
- “findClosestRecursive” function takes an array “arr” , right and left indices of the current search range and integer traget.
- “findClosestRecursive” function usese binary search approach to recursively search array.
- “findClosestRecursive” function also calculates the middle index of the current search range and recursively searches the left and right halves of the array.
- Base case – It occurs when there is only one element in an array. In that case, function will simply returns that element.
C++
#include <bits/stdc++.h>using namespace std;int findClosestRecursive(int arr[], int left, int right, int target) { // base case: when there is only one element in the array if (left == right) { return arr[left]; } // calculate the middle index int mid = (left + right) / 2; // recursively search the left half of the array int leftClosest = findClosestRecursive(arr, left, mid, target); // recursively search the right half of the array int rightClosest = findClosestRecursive(arr, mid + 1, right, target); // compare the absolute differences of the closest elements in the left and right halves if (abs(leftClosest - target) <= abs(rightClosest - target)) { return leftClosest; } else { return rightClosest; }}int main() { int arr[] = {1, 2, 4, 5, 6, 6, 8, 8, 9}; int n = sizeof(arr) / sizeof(arr[0]); int target = 11; cout << findClosestRecursive(arr, 0, n-1, target); return 0;} |
Java
import java.util.*;public class GFG { public static int findClosestRecursive(int[] arr, int left, int right, int target) { // base case: when there is only one element in the // array if (left == right) { return arr[left]; } // calculate the middle index int mid = (left + right) / 2; // recursively search the left half of the array int leftClosest = findClosestRecursive(arr, left, mid, target); // recursively search the right half of the array int rightClosest = findClosestRecursive( arr, mid + 1, right, target); // compare the absolute differences of the closest // elements in the left and right halves if (Math.abs(leftClosest - target) <= Math.abs(rightClosest - target)) { return leftClosest; } else { return rightClosest; } } public static void main(String[] args) { int[] arr = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = arr.length; int target = 11; System.out.println( findClosestRecursive(arr, 0, n - 1, target)); }} |
Python
#Python Code to Find closest number in Sorted array using recursiondef find_closest_recursive(arr, left, right, target): # base case: when there is only one element in the array if left == right: return arr[left] # calculate the middle index mid = (left + right) // 2 # recursively search the left half of the array left_closest = find_closest_recursive(arr, left, mid, target) # recursively search the right half of the array right_closest = find_closest_recursive(arr, mid + 1, right, target) # compare the absolute differences of the closest elements in the left and right halves if abs(left_closest - target) <= abs(right_closest - target): return left_closest else: return right_closestif __name__ == "__main__": arr = [1, 2, 4, 5, 6, 6, 8, 8, 9] n = len(arr) target = 11 print(find_closest_recursive(arr, 0, n-1, target))#This code is contributed by Veerendra_Singh_Rajpoot |
C#
using System;public class GFG{ static int FindClosestRecursive(int[] arr, int left, int right, int target) { // base case: when there is only one element in the array if (left == right) { return arr[left]; } // calculate the middle index int mid = (left + right) / 2; // recursively search the left half of the array int leftClosest = FindClosestRecursive(arr, left, mid, target); // recursively search the right half of the array int rightClosest = FindClosestRecursive(arr, mid + 1, right, target); // compare the absolute differences of the closest elements in the // left and right halves if (Math.Abs(leftClosest - target) <= Math.Abs(rightClosest - target)) { return leftClosest; } else { return rightClosest; } } static void Main(string[] args) { int[] arr = { 1, 2, 4, 5, 6, 6, 8, 8, 9 }; int n = arr.Length; int target = 11; Console.WriteLine(FindClosestRecursive(arr, 0, n - 1, target)); }} |
Javascript
function findClosestRecursive(arr, left, right, target) { // base case: when there is only one element in the array if (left == right) { return arr[left]; } // calculate the middle index let mid = Math.floor((left + right) / 2); // recursively search the left half of the array let leftClosest = findClosestRecursive(arr, left, mid, target); // recursively search the right half of the array let rightClosest = findClosestRecursive(arr, mid + 1, right, target); // compare the absolute differences of the closest elements in the // left and right halves if (Math.abs(leftClosest - target) <= Math.abs(rightClosest - target)) { return leftClosest; } else { return rightClosest; }}let arr = [1, 2, 4, 5, 6, 6, 8, 8, 9];let n = arr.length;let target = 11;console.log(findClosestRecursive(arr, 0, n - 1, target));// This code is contributed by user_dtewbxkn77n |
Output
9
Time Complexity: O(log n)
Auxiliary Space: O(log n)
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