longest consecutive subsequence code example

Example 1: python longest consecutive sequence

def longest(seq = input("""Please enter a series of random letters,
i will guess the largest sequence of letters in order""").upper()):
    
    max_count = 0
    max_char = ""
    prev_char = ""
    
    for current in seq:
        if prev_char == current:
            count += 1
        else:
            count = 1
        if count > max_count:
            max_count = count
            max_char = current
        prev_char = current
    print(max_char,max_count)

longest()

Example 2: longest common subsequence

class Solution:
    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        """
        text1: horizontally
        text2: vertically
        """
        dp = [[0 for _ in range(len(text1)+1)] for _ in range(len(text2)+1)]
        
        for row in range(1, len(text2)+1):
            for col in range(1, len(text1)+1):
                if text2[row-1]==text1[col-1]:
                    dp[row][col] = 1+ dp[row-1][col-1]
                else:
                    dp[row][col] = max(dp[row-1][col], dp[row][col-1])
        return dp[len(text2)][len(text1)]

Example 3: longest common subsequence

int maxSubsequenceSubstring(char x[], char y[], 
                            int n, int m) 
{ 
    int dp[MAX][MAX]; 
  
    // Initialize the dp[][] to 0. 
    for (int i = 0; i <= m; i++) 
        for (int j = 0; j <= n; j++) 
            dp[i][j] = 0; 
  
    // Calculating value for each element. 
    for (int i = 1; i <= m; i++) { 
        for (int j = 1; j <= n; j++) { 
  
            // If alphabet of string X and Y are 
            // equal make dp[i][j] = 1 + dp[i-1][j-1] 
            if (x[j - 1] == y[i - 1]) 
                dp[i][j] = 1 + dp[i - 1][j - 1]; 
  
            // Else copy the previous value in the 
            // row i.e dp[i-1][j-1] 
            else
                dp[i][j] = dp[i][j - 1]; 
        } 
    } 
  
    // Finding the maximum length. 
    int ans = 0; 
    for (int i = 1; i <= m; i++) 
        ans = max(ans, dp[i][n]); 
  
    return ans; 
}