Java Equivalent of C++’s upper_bound() Method - GeeksforGeeks

28 Jan, 2022
In this article, we will discuss Java’s equivalent implementation of the upper_bound() method of C++.  This method is provided with a key-value which is searched in the array. It returns the index of the first element in the array which has a value greater than key or last if no such element is found. The below implementations will find the upper bound value and its index, otherwise, it will print upper bound does not exist.
Illustrations:
Input  : 10 20 30 30 40 50
Output : upper_bound for element 30 is 40 at index 4
Input  : 10 20 30 40 50
Output : upper_bound for element 45 is 50 at index 4
Input  : 10 20 30 40 50
Output : upper_bound for element 60 does not exists
Now let us discuss out the methods in order to use the upper bound() method in order to get the index of the next greater value.
Methods:
Naive Approach (linear search)Iterative binary searchRecursive binary searchbinarySearch() method of Arrays utility class
Naive Approach (linear search)
Iterative binary search
Recursive binary search
binarySearch() method of Arrays utility class
Let us discuss each of the above methods to detailed understanding by providing clean java programs for them as follows:
Method 1: Using linear search  
To find the upper bound using linear search, we will iterate over the array starting from the 0th index until we find a value greater than the key.
Example
Java
// Java program for finding upper bound// using linear search  // Importing Arrays utility classimport java.util.Arrays;  // Main classclass GFG {      // Method 1    // To find upper bound of given key    static void upper_bound(int arr[], int key)    {        int upperBound = 0;          while (upperBound < arr.length) {            // If current value is lesser than or equal to            // key            if (arr[upperBound] <= key)                upperBound++;              // This value is just greater than key            else{                System.out.print("The upper bound of " + key + " is " + arr[upperBound] + " at index " + upperBound);                  return;            }            }        System.out.print("The upper bound of " + key + " does not exist.");    }      // Method 2    // Main driver method    public static void main(String[] args)    {        // Custom array input over which upper bound is to        // be operated by passing a key        int array[] = { 10, 20, 30, 30, 40, 50 };        int key = 30;          // Sort the array using Arrays.sort() method        Arrays.sort(array);          // Printing the upper bound        upper_bound(array, key);    }}
The upper bound of 30 is 40 at index 4
Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(1)
To find the upper bound of a key, we will search the key in the array. 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 examples.  
Method  2: Using binary search iteratively
Procedure:
Sort the array before applying binary search.Initialize low as 0 and high as N.Find the index of the middle element (mid)Compare key with the middle element(arr[mid])If the middle element is less than or equals to key then update the low as mid+1, Else update high as mid.Repeat step 2 to step 4 until low is less than high.After all the above steps the low is the upper_bound of the key
Sort the array before applying binary search.
Initialize low as 0 and high as N.
Find the index of the middle element (mid)
Compare key with the middle element(arr[mid])
If the middle element is less than or equals to key then update the low as mid+1, Else update high as mid.
Repeat step 2 to step 4 until low is less than high.
After all the above steps the low is the upper_bound of the key
Example
Java
// Java program to Find upper bound// Using Binary Search Iteratively  // Importing Arrays utility classimport java.util.Arrays;  // Main classpublic class GFG {      // Iterative approach to find upper bound    // using binary search technique    static void upper_bound(int arr[], int key)    {        int mid, N = arr.length;          // Initialise starting index and        // ending index        int low = 0;        int high = N;          // Till low is less than high        while (low < high && low != N) {            // Find the index of the middle element            mid = low + (high - low) / 2;              // If key is greater than or equal            // to arr[mid], then find in            // right subarray            if (key >= arr[mid]) {                low = mid + 1;            }              // If key is less than arr[mid]            // then find in left subarray            else {                high = mid;            }        }          // If key is greater than last element which is        // array[n-1] then upper bound        // does not exists in the array        if (low == N ) {            System.out.print("The upper bound of " + key + " does not exist.");             return;              }            // Print the upper_bound index          System.out.print("The upper bound of " + key + " is " + arr[low] + " at index " + low);    }      // Driver main method    public static void main(String[] args)    {        int array[] = { 10, 20, 30, 30, 40, 50 };        int key = 30;          // Sort the array using Arrays.sort() method        Arrays.sort(array);          // Printing the upper bound        upper_bound(array, key);    }}
The upper bound of 30 is 40 at index 4
A recursive approach following the same procedure is discussed below:
Method 3: Recursive binary search
Example
Java
// Java program to Find Upper Bound// Using Binary Search Recursively  // Importing Arrays utility classimport java.util.Arrays;  // Main classpublic class GFG {      // Recursive approach to find upper bound    // using binary search technique    static int recursive_upper_bound(int arr[], int low,                                     int high, int key)    {        // Base Case        if (low > high || low == arr.length)            return low;          // Find the value of middle index        int mid = low + (high - low) / 2;          // If key is greater than or equal        // to array[mid], then find in        // right subarray        if (key >= arr[mid]) {            return recursive_upper_bound(arr, mid + 1, high,                                         key);        }          // If key is less than array[mid],        // then find in left subarray        return recursive_upper_bound(arr, low, mid - 1,                                     key);    }      // Method to find upper bound    static void upper_bound(int arr[], int key)    {        // Initialize starting index and        // ending index        int low = 0;        int high = arr.length;          // Call recursive upper bound method        int upperBound            = recursive_upper_bound(arr, low, high, key);        if (upperBound == arr.length)             // upper bound of the key does not exists            System.out.print("The upper bound of " + key                             + " does not exist.");        else System.out.print(                "The upper bound of " + key + " is "                + arr[upperBound] + " at index "                + upperBound);    }      // Main driver method    public static void main(String[] args)    {          int array[] = { 10, 20, 30, 30, 40, 50 };        int key = 30;          // Sorting the array using Arrays.sort() method        Arrays.sort(array);          // Printing the upper bound        upper_bound(array, key);    }}
The upper bound of 30 is 40 at index 4
Method 4: Using binarySearch() method of Arrays utility class
We can also use the in-built binary search method of Arrays utility class (or Collections utility class). This function returns an index of the key in the array. If the key is present in the array it will return its index (not guaranteed to be the first index), otherwise based on sorted order, it will return the expected position of the key i.e (-(insertion point) – 1). 
Approach:
Sort the array before applying binary search.Search the index of the key in a sorted array, if it is present in the array, its index is returned as positive value of , otherwise, a negative value which specifies the position at which the key should be added in the sorted array.Now if the key is present in the array we move rightwards to find next greater value.Print the upper bound index if present.
Sort the array before applying binary search.
Search the index of the key in a sorted array, if it is present in the array, its index is returned as positive value of , otherwise, a negative value which specifies the position at which the key should be added in the sorted array.
Now if the key is present in the array we move rightwards to find next greater value.
Print the upper bound index if present.
Example
Java
// Java program to find upper bound// Using binarySearch() method of Arrays class  // Importing Arrays utility classimport java.util.Arrays;  // Main classpublic class GFG {      // Method 1    // To find upper bound using binary search    // implementation of Arrays utility class    static void upper_bound(int arr[], int key)    {        int index = Arrays.binarySearch(arr, key);        int n = arr.length;          // 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 upper bound            int upperBound = Math.abs(index) - 1;            if (upperBound < n)                System.out.print("The upper bound of " + key                                 + " is " + arr[upperBound]                                 + " at index "                                 + upperBound);            else                System.out.print("The upper bound of " + key                                 + " does not exists.");            return;        }          // If key is present in the array        // we move rightwards to find next greater value        else {              // Increment the index until value is equal to            // key              while (index < n) {                  // If current value is same                if (arr[index] == key)                    index++;                  // Current value is different which means                // it is the greater than the key                else {                    System.out.print(                        "The upper bound of " + key + " is "                        + arr[index] + " at index "                        + index);                    return;                }            }            System.out.print("The upper bound of " + key                             + " does not exist.");        }    }      // Method 2    // Main driver method    public static void main(String[] args)    {        int array[] = { 10, 20, 30, 30, 40, 50 };        int key = 30;          // Sort the array before applying binary search        Arrays.sort(array);          // Printing the lower bound        upper_bound(array, key);    }}
The upper bound of 30 is 40 at index 4
Note: We can also find the index of the middle element via any one of them
int mid = (high + low)/ 2;  
int mid = (low + high) >>> 1;
varshagumber28
TrueGeek-2021
TrueGeek
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Comments
Old Comments
Inventory Management with JSON in Python
MongoDB - Replication and Sharding
How to Provide the Static IP to a Docker Container?
How to Install Golang on CentOS 7?
SQL Query to Compare Results With Today's Date
Top 12 Most Used Git Commands For Developers
Understanding Multi-Layer Feed Forward Networks
How to Install Packages in Flutter?
How to Remove the Last Character From a Table in SQL?
How to Get the Device Information in Android?