Java Equivalent of C++’s lower_bound() Method

• Last Updated : 12 Nov, 2021

The lower_bound() method of C++ return the index of the first element in the array which has a value not less than key. This means that the function returns the index of the next smallest number just greater than or equal to that number. If there are multiple values that are equal to the number, lower_bound() returns the index of the first such value.

Examples:

Input  : 4 6 10 12 18 18 20 20 30 45
Output : lower_bound for element 18 at index 4
Input  : 4 6 10 12 16 20 28
Output : lower_bound for element 18 at index 5
Input  : 24 26 40 56
Output : lower_bound for element 18 at index 0
Input  : 4 6 10 12 16 17
Output : lower_bound for element 18 at index 6

Now let us discuss out the methods in order to use lower bound() method in order to get the index of the next smallest number just greater than or equal to that number.

Methods:

1. Naive Approach
2. Using binary search iteratively
3. Using binary search recursively
4. Using binarySearch() method of Arrays utility class

Method 1: Using linear search

Approach:

We can use linear search to find lower bound. We will iterate over the array starting from the 0th index until we find a value equal to or greater than the key.

Example

Java

 // Java program for finding lower bound// using linear search  // Importing Arrays utility classimport java.util.Arrays;  // Main classclass GFG {      // Method 1    // To find lower bound of given key    static int lower(int array[], int key)    {        int lowerBound = 0;          // Traversing the array using length function        while (lowerBound < array.length) {              // If key is lesser than current value            if (key > array[lowerBound])                lowerBound++;              // This is either the first occurrence of key            // or value just greater than key            else                return lowerBound;        }          return lowerBound;    }      // Method 2    // Main driver method    public static void main(String[] args)    {        // Custom array input over which lower bound is to        // be operated by passing a key        int array[]            = { 4, 6, 10, 12, 18, 18, 20, 20, 30, 45 };        int key = 18;          // Sort the array using Arrays.sort() method        Arrays.sort(array);          // Printing the lower bound        System.out.println(lower(array, key));    }}
Output
4

Time Complexity: O(N), where N is the number of elements in the array.

Auxiliary Space: O(1)

We can use an efficient approach of binary search to search the key in the sorted array in O(log2 n) as proposed in the below example

Method  2: Using binary search iteratively

Procedure:

1. Initialize the low as 0 and high as N.
2. Compare key with the middle element(arr[mid])
3. If the middle element is greater than or equals to key then update the high as a middle index(mid).
4. Else update low as mid + 1.
5. Repeat step 2 to step 4 until low is less than high.
6. After all the above steps the low is the lower_bound of a key in the given array.

Example

Java

 // Java program to Find lower bound// Using Binary Search Iteratively  // Importing Arrays utility classimport java.util.Arrays;  // Main classpublic class GFG {      // Method 1    // Iterative approach to find lower bound    // using binary search technique    static int lower_bound(int array[], int key)    {        // Initialize starting index and        // ending index        int low = 0, high = array.length;        int mid;          // Till high does not crosses low        while (low < high) {              // Find the index of the middle element            mid = low + (high - low) / 2;              // If key is less than or equal            // to array[mid], then find in            // left subarray            if (key <= array[mid]) {                high = mid;            }              // If key is greater than array[mid],            // then find in right subarray            else {                  low = mid + 1;            }        }          // If key is greater than last element which is        // array[n-1] then lower bound        // does not exists in the array        if (low < array.length && array[low] < key) {            low++;        }          // Returning the lower_bound index        return low;    }      // Method 2    // Driver main method    public static void main(String[] args)    {          // Custom array and key input over which lower bound        // is computed        int array[]            = { 4, 6, 10, 12, 18, 18, 20, 20, 30, 45 };        int key = 18;          // Sort the array  using Arrays.sort() method        Arrays.sort(array);          // Printing the lower bound        System.out.println(lower_bound(array, key));    }}
Output
4

Now as usual optimizing further away by providing a recursive approach following the same procedure as discussed above.

Method 3: Using binary search recursively

Java

 // Java program to Find Lower Bound// Using Binary Search Recursively  // Importing Arrays utility classimport java.util.Arrays;  // Main classpublic class GFG {      // Method 1    // To find lower bound using binary search technique    static int recursive_lower_bound(int array[], int low,                                     int high, int key)    {        // Base Case        if (low > high) {            return low;        }          // Find the middle index        int mid = low + (high - low) / 2;          // If key is lesser than or equal to        // array[mid] , then search        // in left subarray        if (key <= array[mid]) {              return recursive_lower_bound(array, low,                                         mid - 1, key);        }          // If key is greater than array[mid],        // then find in right subarray        return recursive_lower_bound(array, mid + 1, high,                                     key);    }      // Method 2    // To compute the lower bound    static int lower_bound(int array[], int key)    {        // Initialize starting index and        // ending index        int low = 0, high = array.length;          // Call recursive lower bound method        return recursive_lower_bound(array, low, high, key);    }      // Method 3    // Main driver method    public static void main(String[] args)    {        // Custom array and key over which lower bound is to        // be computed        int array[]            = { 4, 6, 10, 12, 18, 18, 20, 20, 30, 45 };        int key = 18;          // Sorting the array using Arrays.sort() method        Arrays.sort(array);          // Printing the lower bound        System.out.println(lower_bound(array, key));    }}
Output
4

We can also use the in-built binary search implementation of Arrays utility class (or Collections utility class). The function returns an index of the search key, if it is contained in the array; otherwise, (-(insertion point) – 1). The insertion point is defined as the point at which the key would be inserted into the array.

Approach:

1. Search the index of the key in a sorted array, returns index of the key as positive value of it is present in the array, otherwise, a negative value which specifies.
2. The position at which the key should be added in the sorted array.
3. Now if the key is present in the array we move leftwards to find its first occurrence else decrement the index to find the first occurrence of the key.
4. Sort the array before applying binary search
5. Print it

Example

Java

 // Java program to find lower bound// using binarySearch() method of Arrays class  // Importing Arrays utility classimport java.util.Arrays;  // Main classpublic class GFG {      // Method 1    // To find lower bound using binary search    // implementation of Arrays utility class    static int lower_bound(int array[], int key)    {          int index = Arrays.binarySearch(array, key);          // If key is not present in the array        if (index < 0) {              // Index specify the position of the key            // when inserted in the sorted array            // so the element currently present at            // this position will be the lower bound            return Math.abs(index) - 1;        }          // If key is present in the array        // we move leftwards to find its first occurrence        else {              // Decrement the index to find the first            // occurrence of the key              while (index > 0) {                  // If previous value is same                if (array[index - 1] == key)                    index--;                  // Previous value is different which means                // current index is the first occurrence of                // the key                else                    return index;            }              return index;        }    }      // Method 2    // Main driver method    public static void main(String[] args)    {        //        int array[]            = { 4, 6, 10, 12, 18, 18, 20, 20, 30, 45 };        int key = 18;          // Sort the array before applying binary search        Arrays.sort(array);                // Printing the lower bound         System.out.println(lower_bound(array, key));    }}
Output
4

Note: We can also find mid-value via any one of them

int mid = (high + low)/ 2;
int mid = (low + high) >>> 1;
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