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Problem Name: 2264. Largest 3-Same-Digit Number in String
Problem Link: https://leetcode.com/problems/largest-3-same-digit-number-in-string/
Difficulty: Easy
Tag: String
Language: C# | C++
OJ: LeetCode
Algorithm:
1.Initialize an empty string answer to store the result.
2.Initialize a variable sum to -1 to keep track of the maximum sum found.
3.Iterate through the input string num from the second character (index 1) to the second-to-last character.
2.Initialize a variable sum to -1 to keep track of the maximum sum found.
3.Iterate through the input string num from the second character (index 1) to the second-to-last character.
🏈Calculate the sum of the current three consecutive digits (characters) by converting them to integers and adding them.
🏈Check if the current three digits are equal and if the calculated sum is greater than the current maximum sum.
🏈If both conditions are true, update the answer string and sum.
4.After the iteration, answer will contain the substring with the maximum sum of integer values.
5.Return the answer string.
🏈Check if the current three digits are equal and if the calculated sum is greater than the current maximum sum.
🏈If both conditions are true, update the answer string and sum.
4.After the iteration, answer will contain the substring with the maximum sum of integer values.
5.Return the answer string.
Code(C#)
public class Solution
{
public string LargestGoodInteger(string num)
{
string answer = "";
int sum = -1;
for(int i=1; i<num.Length-1; i++)
{
int added = (num[i-1] - '0') + (num[i] - '0') + (num[i+1] - '0');
if (num[i-1] == num[i] && num[i] == num[i+1] && sum < added)
{
answer = "";
answer += num[i-1];
answer += num[i];
answer += num[i+1];
sum = added;
}
}
return answer;
}
}
Code(C++)
class Solution {
public:
string largestGoodInteger(string num) {
string answer = "";
int sum = -1;
for (int i = 1; i < num.length() - 1; i++) {
int added = (num[i - 1] - '0') + (num[i] - '0') + (num[i + 1] - '0');
if (num[i - 1] == num[i] && num[i] == num[i + 1] && sum < added) {
answer = "";
answer += num[i - 1];
answer += num[i];
answer += num[i + 1];
sum = added;
}
}
return answer;
}
};