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Lexicographic rank of a Binary String
In this article, we will explore an intriguing problem that involves binary strings and lexicographic ordering. Our task is to find the lexicographic rank of a given binary string. We'll demonstrate our solution, a popular programming language known for its efficiency and flexibility.
Understanding Lexicographic Ordering
Lexicographic or lexicographical ordering (also known as alphabetical or dictionary ordering) refers to the arrangement of words based on the alphabetical order of their component letters.
Problem Statement
Given a binary string, we need to determine its lexicographic rank among all its permutations. The lexicographic rank of a string is its position in the set of all permutations of that string when they are listed in lexicographic order.
Solution Approach
Our approach involves these key steps −
Initialize Count − Initialize a counter to store the number of ones ('1's) in the binary string.
Rank Calculation − Iterate through the binary string from left to right. If the current character is '1', calculate its rank using the combination formula, decreasing the counter for each subsequent '1'.
Return Result − The result will be the lexicographic rank of the binary string.
Implementation
Example
The following programs outline the solution −
#include <stdio.h>
#include <string.h>
// Function to calculate factorial
int fact(int n) {
int res = 1;
for (int i = 1; i <= n; i++)
res *= i;
return res;
}
// Function to find lexicographic rank of a binary string
int lexRank(const char* str) {
int rank = 1;
int onesCount = 0;
for (int i = 0; str[i] != '\0'; i++) {
if (str[i] == '1') {
onesCount++;
}
}
for (int i = 0; str[i] != '\0'; i++) {
if (str[i] == '1') {
onesCount--;
rank += fact(onesCount);
}
}
return rank;
}
int main() {
const char* str = "110";
int result = lexRank(str);
printf("Lexicographic rank of the binary string: %d\n", result);
return 0;
}
Output
Lexicographic rank of the binary string: 3
#include <bits/stdc++.h>
using namespace std;
// Function to calculate factorial
int fact(int n) {
int res = 1;
for (int i = 1; i <= n; i++)
res *= i;
return res;
}
// Function to find lexicographic rank of a binary string
int lexRank(string str) {
int rank = 1;
int onesCount = count(str.begin(), str.end(), '1');
for (char c : str) {
if (c == '1') {
onesCount--;
rank += fact(onesCount);
}
}
return rank;
}
int main() {
string str = "110";
int result = lexRank(str);
cout << "Lexicographic rank of the binary string: " << result << endl;
return 0;
}
Output
Lexicographic rank of the binary string: 3
public class Main {
// Function to calculate factorial
static int fact(int n) {
int res = 1;
for (int i = 1; i <= n; i++)
res *= i;
return res;
}
// Function to find lexicographic rank of a binary string
static int lexRank(String str) {
int rank = 1;
int onesCount = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == '1') {
onesCount++;
}
}
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == '1') {
onesCount--;
rank += fact(onesCount);
}
}
return rank;
}
public static void main(String[] args) {
String str = "110";
int result = lexRank(str);
System.out.println("Lexicographic rank of the binary string: " + result);
}
}
Output
Lexicographic rank of the binary string: 3
def fact(n):
res = 1
for i in range(1, n+1):
res *= i
return res
def lex_rank(s):
rank = 1
ones_count = s.count('1')
for c in s:
if c == '1':
ones_count -= 1
rank += fact(ones_count)
return rank
if __name__ == "__main__":
str = "110"
result = lex_rank(str)
print("Lexicographic rank of the binary string:", result)
Output
Lexicographic rank of the binary string: 3
Explanation
Consider the binary string −
str = "110"
The permutations of this binary string are: "011", "101", "110". In lexicographic order, these permutations are: "011", "101", "110".
The rank of the binary string "110" is 3, which is the output of our program.
Conclusion
The problem of finding the lexicographic rank of a binary string is a fascinating one that builds upon our understanding of binary strings, permutations, and lexicographic order. This solution, implemented in C, C++, Java and Python demonstrates how we can use basic programming constructs to solve it effectively.
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