# 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:

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 absolution difference.

An efficient solution is to use Binary Search.

## C++

 // CPP program to find element // closet to given target. #include 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 code int 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 closet 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));     } }

## Python 3

 # Python3 program to find element # closet 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 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  // closet 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

= \$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.
?>

Output:

9

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : harishkumar88