Java equivalent of c++ equal_range (or lower_bound & upper_bound)
In Java, you use Collections.binarySearch to find the lower bound of the equal range in a sorted list (Arrays.binarySearch provides a similar capability for arrays). This gives you a position within the equal range with no further guarantees:
If the list contains multiple elements equal to the specified object, there is no guarantee which one will be found.
Then you iterate linearly forward and then backward until you hit the end of the equal range.
These methods work for objects implementing the Comparable interface. For classes that do not implement the Comparable, you can supply an instance of a custom Comparator for comparing the elements of your specific type.
We can find Lower bound and upper bound with the help of java library function as well as by defining our own LowerBound and UpperBound Function.
{#case-1}
if the number is not present both lower bound and upper bound would be same .i.e. in that case lb and ub would be the insertion point of the array i.e. that point where the number should be inserted to keep the array sorted.
Example-1:
6 1 // 6 is the size of the array and 1 is the key
2 3 4 5 6 7 here lb=0 and ub=0 (0 is the position where 1 should be inserted to keep the array sorted)
6 8 // 6 is the size of the array and 8 is the key
2 3 4 5 6 7 here lb=6 and ub=6 (6 is the position where 8 should be inserted to keep the array sorted)
6 3 // 6 is the size of the array and 3 is the key
1 2 2 2 4 5 here lb=4 and ub=4 (4 is the position where 3 should be inserted to keep the array sorted)
{#case-2(a)}
if the number is present and have frequency 1. i.e. number of occurrence is 1
lb=index of that number.
ub=index of the next number which is just greater than that number in the array
.i.e. ub=index of that number+1
Example-2:
6 5 // 6 is the size of the array and 5 is the key
1 2 3 4 5 6 here lb=4 and ub=5
{#case-2(b)}
if the number is present and have frequency more than 1. number is occured multiple times.in this case lb would be the index of the 1st occurrence of that number. ub would be the index of the last occurrence of that number+1. i.e. index of that number which is just greater than the key in the array.
Example-3:
11 5 // 11 is the size of the array and 5 is the key
1 2 3 4 5 5 5 5 5 7 7 here lb=4 and ub=9
Implementation of Lower_Bound and Upper_Bound
Method-1: By Library function
// a is the array and x is the target value
int lb=Arrays.binarySearch(a,x); // for lower_bound
int ub=Arrays.binarySearch(a,x); // for upper_bound
if(lb<0) {lb=Math.abs(lb)-1;}//if the number is not present
else{ // if the number is present we are checking
//whether the number is present multiple times or not
int y=a[lb];
for(int i=lb-1; i>=0; i--){
if(a[i]==y) --lb;
else break;
}
}
if(ub<0) {ub=Math.abs(ub)-1;}//if the number is not present
else{// if the number is present we are checking
//whether the number is present multiple times or not
int y=a[ub];
for(int i=ub+1; i<n; i++){
if(a[i]==y) ++ub;
else break;
}
++ub;
}
Method-2: By Defining own Function
//for lower bound
static int LowerBound(int a[], int x) { // x is the target value or key
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
// for Upper_Bound
static int UpperBound(int a[], int x) {// x is the key or target value
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]<=x) l=m;
else r=m;
}
return l+1;
}
or we can use
int m=l+(r-l)/2;
but if we use
int m=(l+r)>>>1; // it is probably faster
but the usage of any of the above formula of calculating m will prevent overflow
In C and C++ (>>>) operator is absent, we can do this:
int m= ((unsigned int)l + (unsigned int)r)) >> 1;
// implementation in program:
import java.util.*;
import java.lang.*;
import java.io.*;
public class Lower_bound_and_Upper_bound {
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer s = new StringTokenizer(br.readLine());
int n=Integer.parseInt(s.nextToken()),x=Integer.parseInt(s.nextToken()),a[]=new int[n];
s = new StringTokenizer(br.readLine());
for(int i=0; i<n; i++) a[i]=Integer.parseInt(s.nextToken());
Arrays.sort(a);// Array should be sorted. otherwise lb and ub cant be calculated
int u=UpperBound(a,x);
int l=LowerBound(a,x);
System.out.println(l+" "+u);
}
}
# Equivalent C++ code for calculating lowerbound and upperbound
#include<bits/stdc++.h>
#define IRONMAN ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long int ll;
int main() {
IRONMAN
int n,x;cin>>n>>x;
vector<int> v(n);
for(auto &i: v) cin>>i;
ll lb=(lower_bound(v.begin(),v.end(),x))-v.begin();// for calculating lb
ll ub=(upper_bound(v.begin(),v.end(),x))-v.begin();// for calculating ub
cout<<lb<<" "<<ub<<"\n";
return 0;
}
Java have already built-in binary search functionality that calculates lower/upper bounds for an element in an array, there is no need to implement custom methods.
When we speak about upper/lower bounds or equal ranges, we always mean indexes of a container (in this case of ArrayList), and not the elements contained. Let's consider an array (we assume the array is sorted, otherwise we sort it first):
List<Integer> nums = new ArrayList<>(Arrays.asList(2,3,5,5,7,9,10,18,22));
The "lower bound" function must return the index of the array, where the element must be inserted to keep the array sorted. The "upper bound" must return the index of the smallest element in the array, that is bigger than the looked for element. For example
lowerBound(nums, 6)
must return 3, because 3 is the position of the array (starting counting with 0), where 6 must be inserted to keep array sorted.
The
upperBound(nums, 6)
must return 4, because 4 is the position of the smallest element in the array, that is bigger than 5 or 6, (number 7 on position 4).
In C++ in standard library the both algorithms already implemented in standard library. In Java you can use
Collections.binarySearch(nums, element)
to calculate the position in logarithmic time complexity.
If the array contains the element, Collections.binarySearch returns the first index of the element (in the array above 2). Otherwise it returns a negative number that specifies the position in the array of the next bigger element, counting backwards from the last index of the array. The number found in this position is the smallest element of the array that is bigger than the element you look for.
For example, if you call
int idx = Collections.binarySearch(nums, 6)
the function returns -5. If you count backwards from the last index of the array (-1, -2, ...) the index -5 points to number 7 - the smallest number in the array that is bigger than the element 6.
Conclusion: if the sorted array contains the looked for element, the lower bound is the position of the element, and the upper bound is the position of the next bigger element.
If the array does not contains the element, the lower bound is the position
Math.abs(idx) - 2
and the upper bound is the position
Math.abs(idx) - 1
where
idx = Collections.binarySearch(nums, element)
And please always keep in mind the border cases. For example, if you look for 1 in the above specified array:
idx = Collections.binarySearch(nums, 1)
The functon returns -1. So, the upperBound = Math.abs(idx) - 1 = 0 - the element 2 at position 0. But there is no lower bound for element 1, because 2 is the smallest number in the array. The same logic applies to elements bigger than the biggest number in the array: if you look for lower/upper bounds of number 25, you will get
idx = Collections.binarySearch(nums, 25)
ix = -10. You can calculate the lower bound : lb = Math.abs(-10) - 2 = 8, that is the last index of the array, but there is no upper bound, because 22 is already biggest element in the array and there is no element at position 9.
The equal_range specifies all indexes of the array in the range starting from the lower bound index up to (but not including) the upper bound. For example, the equal range of number 5 in the array above are indexes
[2,3]
Equal range of number 6 is empty, because there is no number 6 in the array.