714144. Geek hates too many 1s
Medium
Geek Hates Too Many 1s
Summary
The problem is about finding a binary string that has the least number of ones among all possible binary strings that satisfy certain conditions. This involves bit manipulation and data structures concepts.
Detailed Explanation
To solve this problem, we can start by generating all possible binary strings with a given length n. We can then count the number of ones in each string and find the one with the least number of ones.
Here is a step-by-step breakdown of the solution:
- Generate all possible binary strings: This can be done using bitwise operations to generate the next string in the sequence.
- Count the number of ones in each string: This involves iterating through each character in the string and incrementing a counter whenever we encounter a '1'.
- Find the string with the least number of ones: We can do this by comparing the count of ones for each string and returning the one with the lowest count.
The time complexity of this solution is O(2^n) because we are generating all possible binary strings, where n is the length of the string. The space complexity is also O(2^n) because we need to store all these strings in memory.
Optimized Solutions
Java
public class GeekHatesTooMany1s {
public static void main(String[] args) {
int n = 3; // length of the binary string
String result = findStringWithLeastOnes(n);
System.out.println(result);
}
public static String findStringWithLeastOnes(int n) {
int minOnes = Integer.MAX_VALUE;
String result = "";
for (int i = 0; i < (1 << n); i++) { // generate all possible binary strings
String binStr = Integer.toBinaryString(i);
while (binStr.length() < n) {
binStr = "0" + binStr; // pad with zeros
}
int onesCount = 0;
for (int j = 0; j < n; j++) {
if (binStr.charAt(j) == '1') {
onesCount++;
}
}
if (onesCount < minOnes) {
minOnes = onesCount;
result = binStr;
}
}
return result;
}
}