Binary Search

• Difficulty Level : Easy
• Last Updated : 18 Jan, 2022

Given a sorted array arr[] of n elements, write a function to search a given element x in arr[].
A simple approach is to do a linear search. The time complexity of the above algorithm is O(n). Another approach to perform the same task is using Binary Search.
Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise, narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.

Example : The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Log n).

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

We basically ignore half of the elements just after one comparison.

1. Compare x with the middle element.
2. If x matches with the middle element, we return the mid index.
3. Else If x is greater than the mid element, then x can only lie in the right half subarray after the mid element. So we recur for the right half.
4. Else (x is smaller) recur for the left half.

Recursive implementation of Binary Search

C++

 // C++ program to implement recursive Binary Search#include using namespace std; // A recursive binary search function. It returns// location of x in given array arr[l..r] is present,// otherwise -1int binarySearch(int arr[], int l, int r, int x){    if (r >= l) {        int mid = l + (r - l) / 2;         // If the element is present at the middle        // itself        if (arr[mid] == x)            return mid;         // If element is smaller than mid, then        // it can only be present in left subarray        if (arr[mid] > x)            return binarySearch(arr, l, mid - 1, x);         // Else the element can only be present        // in right subarray        return binarySearch(arr, mid + 1, r, x);    }     // We reach here when element is not    // present in array    return -1;} int main(void){    int arr[] = { 2, 3, 4, 10, 40 };    int x = 10;    int n = sizeof(arr) / sizeof(arr);    int result = binarySearch(arr, 0, n - 1, x);    (result == -1)        ? cout << "Element is not present in array"        : cout << "Element is present at index " << result;    return 0;}

C

 // C program to implement recursive Binary Search#include  // A recursive binary search function. It returns// location of x in given array arr[l..r] is present,// otherwise -1int binarySearch(int arr[], int l, int r, int x){    if (r >= l) {        int mid = l + (r - l) / 2;         // If the element is present at the middle        // itself        if (arr[mid] == x)            return mid;         // If element is smaller than mid, then        // it can only be present in left subarray        if (arr[mid] > x)            return binarySearch(arr, l, mid - 1, x);         // Else the element can only be present        // in right subarray        return binarySearch(arr, mid + 1, r, x);    }     // We reach here when element is not    // present in array    return -1;} int main(void){    int arr[] = { 2, 3, 4, 10, 40 };    int n = sizeof(arr) / sizeof(arr);    int x = 10;    int result = binarySearch(arr, 0, n - 1, x);    (result == -1)        ? printf("Element is not present in array")        : printf("Element is present at index %d", result);    return 0;}

Java

 // Java implementation of recursive Binary Searchclass BinarySearch {    // Returns index of x if it is present in arr[l..    // r], else return -1    int binarySearch(int arr[], int l, int r, int x)    {        if (r >= l) {            int mid = l + (r - l) / 2;             // If the element is present at the            // middle itself            if (arr[mid] == x)                return mid;             // If element is smaller than mid, then            // it can only be present in left subarray            if (arr[mid] > x)                return binarySearch(arr, l, mid - 1, x);             // Else the element can only be present            // in right subarray            return binarySearch(arr, mid + 1, r, x);        }         // We reach here when element is not present        // in array        return -1;    }     // Driver method to test above    public static void main(String args[])    {        BinarySearch ob = new BinarySearch();        int arr[] = { 2, 3, 4, 10, 40 };        int n = arr.length;        int x = 10;        int result = ob.binarySearch(arr, 0, n - 1, x);        if (result == -1)            System.out.println("Element not present");        else            System.out.println("Element found at index "                               + result);    }}/* This code is contributed by Rajat Mishra */

Python3

 # Python3 Program for recursive binary search. # Returns index of x in arr if present, else -1  def binarySearch(arr, l, r, x):     # Check base case    if r >= l:         mid = l + (r - l) // 2         # If element is present at the middle itself        if arr[mid] == x:            return mid         # If element is smaller than mid, then it        # can only be present in left subarray        elif arr[mid] > x:            return binarySearch(arr, l, mid-1, x)         # Else the element can only be present        # in right subarray        else:            return binarySearch(arr, mid + 1, r, x)     else:        # Element is not present in the array        return -1  # Driver Codearr = [2, 3, 4, 10, 40]x = 10 # Function callresult = binarySearch(arr, 0, len(arr)-1, x) if result != -1:    print("Element is present at index % d" % result)else:    print("Element is not present in array")

C#

 // C# implementation of recursive Binary Searchusing System; class GFG {    // Returns index of x if it is present in    // arr[l..r], else return -1    static int binarySearch(int[] arr, int l, int r, int x)    {        if (r >= l) {            int mid = l + (r - l) / 2;             // If the element is present at the            // middle itself            if (arr[mid] == x)                return mid;             // If element is smaller than mid, then            // it can only be present in left subarray            if (arr[mid] > x)                return binarySearch(arr, l, mid - 1, x);             // Else the element can only be present            // in right subarray            return binarySearch(arr, mid + 1, r, x);        }         // We reach here when element is not present        // in array        return -1;    }     // Driver method to test above    public static void Main()    {         int[] arr = { 2, 3, 4, 10, 40 };        int n = arr.Length;        int x = 10;         int result = binarySearch(arr, 0, n - 1, x);         if (result == -1)            Console.WriteLine("Element not present");        else            Console.WriteLine("Element found at index "                              + result);    }} // This code is contributed by Sam007.

PHP

 = \$l){        \$mid = ceil(\$l + (\$r - \$l) / 2);         // If the element is present        // at the middle itself        if (\$arr[\$mid] == \$x)            return floor(\$mid);         // If element is smaller than        // mid, then it can only be        // present in left subarray        if (\$arr[\$mid] > \$x)            return binarySearch(\$arr, \$l,                                \$mid - 1, \$x);         // Else the element can only        // be present in right subarray        return binarySearch(\$arr, \$mid + 1,                            \$r, \$x);} // We reach here when element// is not present in arrayreturn -1;} // Driver Code\$arr = array(2, 3, 4, 10, 40);\$n = count(\$arr);\$x = 10;\$result = binarySearch(\$arr, 0, \$n - 1, \$x);if((\$result == -1))echo "Element is not present in array";elseecho "Element is present at index ",                            \$result;                             // This code is contributed by anuj_67.?>

Javascript



Output :

Element is present at index 3

Here you can create a check function for easier implementation.

Here is recursive implementation with check function which I feel is a much easier implementation:

C++

 #include using namespace std; // define array globallyconst int N = 1e6 + 4; int a[N];int n; // array size // element to be searched in arrayint k; bool check(int dig){    // element at dig position in array    int ele = a[dig];     // if k is less than    // element at dig position    // then we need to bring our higher ending to dig    // and then continue further    if (k <= ele) {        return 1;    }    else {        return 0;    }}void binsrch(int lo, int hi){    while (lo < hi) {        int mid = (lo + hi) / 2;        if (check(mid)) {            hi = mid;        }        else {            lo = mid + 1;        }    }    // if a[lo] is k    if (a[lo] == k)        cout << "Element found at index "             << lo; // 0 based indexing    else        cout            << "Element doesnt exist in array"; // element                                                // was not in                                                // our array} int main(){    cin >> n;    for (int i = 0; i < n; i++) {        cin >> a[i];    }    cin >> k;     // it is being given array is sorted    // if not then we have to sort it     // minimum possible point where our k can be is starting    // index so lo=0 also k cannot be outside of array so end    // point hi=n     binsrch(0, n);     return 0;}

Iterative implementation of Binary Search

C++

 // C++ program to implement recursive Binary Search#include using namespace std; // A iterative binary search function. It returns// location of x in given array arr[l..r] if present,// otherwise -1int binarySearch(int arr[], int l, int r, int x){    while (l <= r) {        int m = l + (r - l) / 2;         // Check if x is present at mid        if (arr[m] == x)            return m;         // If x greater, ignore left half        if (arr[m] < x)            l = m + 1;         // If x is smaller, ignore right half        else            r = m - 1;    }     // if we reach here, then element was    // not present    return -1;} int main(void){    int arr[] = { 2, 3, 4, 10, 40 };    int x = 10;    int n = sizeof(arr) / sizeof(arr);    int result = binarySearch(arr, 0, n - 1, x);    (result == -1) ? cout << "Element is not present in array"                   : cout << "Element is present at index " << result;    return 0;}

C

 // C program to implement iterative Binary Search#include  // A iterative binary search function. It returns// location of x in given array arr[l..r] if present,// otherwise -1int binarySearch(int arr[], int l, int r, int x){    while (l <= r) {        int m = l + (r - l) / 2;         // Check if x is present at mid        if (arr[m] == x)            return m;         // If x greater, ignore left half        if (arr[m] < x)            l = m + 1;         // If x is smaller, ignore right half        else            r = m - 1;    }     // if we reach here, then element was    // not present    return -1;} int main(void){    int arr[] = { 2, 3, 4, 10, 40 };    int n = sizeof(arr) / sizeof(arr);    int x = 10;    int result = binarySearch(arr, 0, n - 1, x);    (result == -1) ? printf("Element is not present"                            " in array")                   : printf("Element is present at "                            "index %d",                            result);    return 0;}

Java

 // Java implementation of iterative Binary Searchclass BinarySearch {    // Returns index of x if it is present in arr[],    // else return -1    int binarySearch(int arr[], int x)    {        int l = 0, r = arr.length - 1;        while (l <= r) {            int m = l + (r - l) / 2;             // Check if x is present at mid            if (arr[m] == x)                return m;             // If x greater, ignore left half            if (arr[m] < x)                l = m + 1;             // If x is smaller, ignore right half            else                r = m - 1;        }         // if we reach here, then element was        // not present        return -1;    }     // Driver method to test above    public static void main(String args[])    {        BinarySearch ob = new BinarySearch();        int arr[] = { 2, 3, 4, 10, 40 };        int n = arr.length;        int x = 10;        int result = ob.binarySearch(arr, x);        if (result == -1)            System.out.println("Element not present");        else            System.out.println("Element found at "                               + "index " + result);    }}

Python3

 # Python3 code to implement iterative Binary# Search. # It returns location of x in given array arr# if present, else returns -1def binarySearch(arr, l, r, x):     while l <= r:         mid = l + (r - l) // 2;                 # Check if x is present at mid        if arr[mid] == x:            return mid         # If x is greater, ignore left half        elif arr[mid] < x:            l = mid + 1         # If x is smaller, ignore right half        else:            r = mid - 1         # If we reach here, then the element    # was not present    return -1 # Driver Codearr = [ 2, 3, 4, 10, 40 ]x = 10 # Function callresult = binarySearch(arr, 0, len(arr)-1, x) if result != -1:    print ("Element is present at index % d" % result)else:    print ("Element is not present in array")

C#

 // C# implementation of iterative Binary Searchusing System; class GFG {    // Returns index of x if it is present in arr[],    // else return -1    static int binarySearch(int[] arr, int x)    {        int l = 0, r = arr.Length - 1;        while (l <= r) {            int m = l + (r - l) / 2;             // Check if x is present at mid            if (arr[m] == x)                return m;             // If x greater, ignore left half            if (arr[m] < x)                l = m + 1;             // If x is smaller, ignore right half            else                r = m - 1;        }         // if we reach here, then element was        // not present        return -1;    }     // Driver method to test above    public static void Main()    {        int[] arr = { 2, 3, 4, 10, 40 };        int n = arr.Length;        int x = 10;        int result = binarySearch(arr, x);        if (result == -1)            Console.WriteLine("Element not present");        else            Console.WriteLine("Element found at "                              + "index " + result);    }}// This code is contributed by Sam007



Javascript



Output :

Element is present at index 3

“Bitwise binary search”

Idea:

Every number can be represented as a sum of the powers of the number 2.

Examples:

1. 76 = 64 + 8 + 4
2. 10 = 8  + 2
3. 7 = 4 + 2 + 1

Approach:

Compute the first power of 2 that is greater or equal then the size of the array.

Initialize an index as 0.

Loop while the computed power is greater than 0 and each time divide it by 2.

Each time the element at position [index + power] <= target we add to the index variable the respective power value. (Build the sum)

After the for loops check if the element at position [index] == target. If so the target element is present in the array, else not.

(no division needed only addition and bitwise shifting)

C++

 // C++ program to implement Bitwise Binary Search #include using namespace std; int binary_search(int *arr,int size,int target){    int index, power;       //Compute the first power of 2 that is >= size    for (power = 1; power < size; power <<= 1);       //loop while(power > 0)      //and divide power by two each iteration    for (index = 0; power; power >>= 1){                 //if the next condition is true          //it means that the power value can contribute to the "sum"(a closer index where target might be)        if (index + power < size && arr[index + power] <= target)           index += power;            }       //if the element at position [index] == target,      //the target value is present in the array    if(arr[index] == target)        return index;         //else the value is not present in the array    return -1;} int main(){    int arr = {1, 3, 5, 7, 8};    int size = 5;    int x = 3;    int answer = binary_search(arr, size, x);    if(answer == -1)        cout << "Element not found";    else        cout << "Element found at position " << answer;}  // This code is contributed  // by Gatea David

Java

 // Java program to implement Bitwise Binary Searchimport java.util.*; class GFG{   static int binary_search(int []arr,int size,int target)  {    int index, power;     // Compute the first power of 2 that is >= size    for (power = 1; power < size; power <<= 1);     // loop while(power > 0)    // and divide power by two each iteration    for (index = 0; power>0; power >>= 1){       // if the next condition is true      // it means that the power value can      // contribute to the "sum"(a closer index where target might be)      if (index + power < size && arr[index + power] <= target)        index += power;      }     // if the element at position [index] == target,    // the target value is present in the array    if(arr[index] == target)      return index;     // else the value is not present in the array    return -1;  }   // Driver code  public static void main(String[] args)  {    int []arr = {1, 3, 5, 7, 8};    int size = 5;    int x = 3;    int answer = binary_search(arr, size, x);    if(answer == -1)      System.out.print("Element not found");    else      System.out.print("Element found at position " +  answer);  }} // This code is contributed by Rajput-Ji

Time Complexity:

The time complexity of Binary Search can be written as:

T(n) = T(n/2) + c

The above recurrence can be solved either using the Recurrence Tree method or the Master method. It falls in case II of the Master Method and the solution of the recurrence is .
Auxiliary Space: O(1) in case of iterative implementation. In the case of recursive implementation, O(Logn) recursion call stack space.

Note:

Here we are using

int mid = low + (high – low)/2;

Maybe, you wonder why we are calculating the middle index this way, we can simply add the lower and higher index and divide it by 2.

int mid = (low + high)/2;

But if we calculate the middle index like this means our code is not 100% correct, it contains bugs.

That is, it fails for larger values of int variables low and high. Specifically, it fails if the sum of low and high is greater than the maximum positive int value(231 – 1 ).

The sum overflows to a negative value and the value stays negative when divided by 2. In java, it throws ArrayIndexOutOfBoundException.

int mid = low + (high – low)/2;

So it’s better to use it like this. This bug applies equally to merge sort and other divide and conquer algorithms.

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