Alternating Character Substrings — Problem Statement & Solution Guide
Problem Description
Given a string mealCodes consisting of only two distinct characters, 'V' and 'N', find the total number of substrings where the characters alternate between 'V' and 'N'.
Examples
Input
VN
Output
3
Explanation: Step-by-step: For input 'VN', we count the substrings 'V', 'VN', and 'N'. The substring 'V' is counted because it's a single character, 'VN' is counted because it's an alternating substring, and 'N' is not counted because it's a single character. Therefore, the total count is 3.
Input
NVVN
Output
6
Explanation: Step-by-step: For input 'NVVN', we count the substrings 'N', 'NV', 'V', 'VN', 'N', and 'NV'. The substrings 'N', 'V', and 'N' are counted because they're single characters, 'NV' and 'VN' are counted because they're alternating substrings, and 'V' is not counted because it's a single character. Therefore, the total count is 6.
Constraints
- 1 <= length of string <= 1000
- The string will only contain the characters 'V' and 'N'.
Optimal Approach & Strategy
A more efficient approach involves iterating through the string once and using two pointers to track the start and end of potential substrings. This approach would have a time complexity of O(n) and a space complexity of O(1), making it more efficient for large inputs.
Brute Force Approach
A brute-force approach would involve generating all possible substrings of the input string and checking each one to see if it alternates between 'V' and 'N'. This would result in a time complexity of O(n^2) due to the nested loops. Additionally, it would require extra space to store the substrings.
Verified Code Solutions
function countAlternatingSubstrings(mealCodes) {
let count = 0;
for (let i = 0; i < mealCodes.length; i++) {
let j = i + 1;
while (j < mealCodes.length) {
if (mealCodes[i] === mealCodes[j]) {
j++;
} else if (j - i > 1 && mealCodes[i] === mealCodes[j - 1]) {
j++;
} else {
count++;
j = i + 1;
}
}
}
return count;
}class Solution {
public int solution(String mealCodes) {
int count = 0;
for (int i = 0; i < mealCodes.length(); i++) {
for (int j = i + 1; j <= mealCodes.length(); j++) {
String substring = mealCodes.substring(i, j);
if (substring.chars().distinct().count() == 2 && substring.contains("V") && substring.contains("N")) {
if ((substring.charAt(0) == 'V' && substring.charAt(1) == 'N') || (substring.charAt(0) == 'N' && substring.charAt(1) == 'V')) {
count++;
}
}
}
}
return count;
}
}def solution(mealCodes: str) -> int:
count = 0
for i in range(len(mealCodes)):
for j in range(i + 1, len(mealCodes) + 1):
substring = mealCodes[i:j]
if len(set(substring)) == 2 and 'V' in substring and 'N' in substring:
if substring[0] == 'V' and substring[1] == 'N' or substring[0] == 'N' and substring[1] == 'V':
count += 1
return countfunction countAlternatingSubstrings(mealCodes) {
let count = 0;
for (let i = 0; i < mealCodes.length; i++) {
let j = i + 1;
while (j < mealCodes.length) {
if (mealCodes[i] === mealCodes[j]) {
j++;
} else if (j - i > 1 && mealCodes[i] === mealCodes[j - 1]) {
j++;
} else {
count++;
j = i + 1;
}
}
}
return count;
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