LeetCode 1529: Minimum Suffix Flips Solution
Master LeetCode problem 1529 (Minimum Suffix Flips), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation:
To solve this problem, we can iterate from right to left over the target string. If the current character in the target string is different from the corresponding character in the string s, we need to flip the range from the current index to the end of the string.
We start from the right end because if we start from the left and flip a range, it might affect the bits to the right of the current index which we have not processed yet.
After processing all the characters in the target string, we will have the minimum number of operations needed to make s equal to the target string.
Time Complexity:
The time complexity of this approach is O(n) where n is the length of the target string.
Space Complexity:
The space complexity is O(1) as we are not using any extra space apart from the input strings.
:
Solution Code
class Solution {
public int minFlips(String target) {
int flips = 0;
char prev = '0';
for (int i = 0; i < target.length(); i++) {
if (target.charAt(i) != prev) {
flips++;
prev = target.charAt(i);
}
}
return flips;
}
}Try It Yourself
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What is the time complexity of LeetCode 1529 (Minimum Suffix Flips)?
The time complexity for LeetCode 1529 (Minimum Suffix Flips) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.
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