Solutions for String pattern matching

    • @Geraldpof
    • @Charles, submitted in Python 3, achived 100% score
    # Complete the substrCount function below.
    def solve(n, s):
        count = n
        for x in range(n): 
            y = x
            while y < n - 1:
                y += 1
                if s[x] == s[y]:
                    count += 1
                else:
                    if s[x:y] == s[y+1:2*y-x+1]:
                        count += 1
                    break
    
        return count
    
    • @
    • @Zafir, submitted in Python 2, achived 100% score
    # Complete the substrCount function below.
    def solve(n, s):
        count = n
        for x in range(n): 
            y = x
            while y < n - 1:
                y += 1
                if s[x] == s[y]:
                    count += 1
                else:
                    if s[x:y] == s[y+1:2*y-x+1]:
                        count += 1
                    break
    
        return count
    
    
    • @
    • @Zafir, submitted in Python 2, achived 100% score
    # Complete the substrCount function below.
    def solve(n, s):
        count = n
        for x in range(n): 
            y = x
            while y < n - 1:
                y += 1
                if s[x] == s[y]:
                    count += 1
                else:
                    if s[x:y] == s[y+1:2*y-x+1]:
                        count += 1
                    break
    
        return count
    
    
    • @JavierVow
    • @jacob, submitted in Python 2, achived 100% score
    # Complete the substrCount function below.
    def solve(n, s):
        count = n
        for x in range(n): 
            y = x
            while y < n - 1:
                y += 1
                if s[x] == s[y]:
                    count += 1
                else:
                    if s[x:y] == s[y+1:2*y-x+1]:
                        count += 1
                    break
    
        return count
    
    
    • @JavierVow
    • @jacob, submitted in Python 3, achived 100% score
    # Complete the substrCount function below.
    def solve(n, s):
        count = n
        for x in range(n): 
            y = x
            while y < n - 1:
                y += 1
                if s[x] == s[y]:
                    count += 1
                else:
                    if s[x:y] == s[y+1:2*y-x+1]:
                        count += 1
                    break
    
        return count