Find closest number in array
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:
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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 : 5A 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 absolution difference.
An efficient solution is to use Binary Search.
C++
// CPP program to find element// closest to given target.#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 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); // 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, 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.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.# 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 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. |
PHP
<?php// PHP program to find element closest// to given target.// 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 bu Sachin.?> |
Javascript
<script>// JavaScript program to find element// closest to given target.// 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> |
Output:
9
