Find closest number in Sorted 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:

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 code
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));
}

// 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 code
arr = [1, 2, 4, 5, 6, 6, 8, 9]
n = len(arr)
target = 11
print(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.

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 bu Sachin.
?>

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 Code
let 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

Time Complexity: O(log  n) (Due to Binary Search)
Auxiliary Space: O(log n) (implicit stack is created due to recursion)