Last updated on October 9th, 2024 at 05:41 pm

Here, We see Student Attendance Record II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Student Attendance Record II LeetCode Solution

Problem Statement

An attendance record for a student can be represented as a string where each character signifies whether the student was absent, late, or present on that day. The record only contains the following three characters:

• 'A': Absent.
• 'L': Late.
• 'P': Present.

Any student is eligible for an attendance award if they meet both of the following criteria:

• The student was absent ('A') for strictly fewer than 2 days total.
• The student was never late ('L') for 3 or more consecutive days.

Given an integer n, return the number of possible attendance records of length n that make a student eligible for an attendance award. The answer may be very large, so return it modulo 109 + 7.

Example 1:
Input: n = 2
Output: 8
Explanation: There are 8 records with length 2 that are eligible for an award: “PP”, “AP”, “PA”, “LP”, “PL”, “AL”, “LA”, “LL” Only “AA” is not eligible because there are 2 absences (there need to be fewer than 2).

Example 2:
Input: n = 1
Output: 3

Example 3:
Input: n = 10101
Output: 183236316

1. Student Attendance Record II LeetCode Solution C++

class Solution {
public:
    int checkRecord(int n) {
long i;
long a0, a1, a2, a3;
long p0, p1, p2;
long l0, l1, l2;
ulong m = 1000000007;
if(n == 1) return 3;
if(n == 2) return 8;
a3 = 1;
a2 = 1;
a1 = 2;
p2 = 1;
p1 = 3;
l1 = 3;
for(i=3; i<=n; i++){
    p0 = (a1 + p1 + l1)%m;
    a0 = (a1 + a2 + a3)%m;
    l0 = (p1 + a1 + p2 + a2)%m;
    p2 = p1%m;
    p1 = p0%m;
    a3 = a2%m;
    a2 = a1%m;
    a1 = a0%m;
    l1 = l0%m;
}
return (a0 + p0 + l0)%m;        
    }
};

2. Student Attendance Record II LeetCode Solution Java

class Solution {
    private int mod= 1000000007;
    int dp[][][] = new int[100002][2][3];
    public int solve(int ind,int n,int cnt_abs,int cnt_late)
    {
        if(ind >= n)
        return 1;
        if(dp[ind][cnt_abs][cnt_late]!=-1)
        return dp[ind][cnt_abs][cnt_late];
            int abs= cnt_abs==0?solve(ind+1,n,1,0):0;
            int pre = solve(ind+1,n,cnt_abs,0);
            int late=0;
            if(cnt_late!=2)
            late=solve(ind+1,n,cnt_abs,cnt_late+1);
            return dp[ind][cnt_abs][cnt_late]=((pre+late)%mod+abs)%mod;
    }
    public int checkRecord(int n) {
         for (int i = 0; i < 100002; i++) {
             for (int j = 0; j < 2; j++) {
                 for (int k = 0; k < 3; k++) {
                     dp[i][j][k] = -1;
        }
    }
}
              return solve(0,n,0,0);
    }
}

3. Student Attendance Record II LeetCode Solution JavaScript

let mod = 1e9 + 7;
function fn(n, i, j, res) {
    if (n === 0) return 1;
    if (res[n][i][j] !== mod) return res[n][i][j];
    let val1 = 0, val2 = 0, val3 = 0;
    if (i > 0) val1 = fn(n - 1, i - 1, 2, res) % mod;
    if (j > 0) val2 = fn(n - 1, i, j - 1, res) % mod;
    val3 = fn(n - 1, i, 2, res) % mod;
    return res[n][i][j] = ((val1 % mod + val2 % mod) % mod + val3 % mod) % mod;
}

var checkRecord = function (n) {
    let i = 1, j = 2, res = [];
    for (let i = 0; i <= n; i++) {
        let arr = [];
        for (let i = 0; i <= 1; i++) arr.push(new Array(3).fill(mod));
        res.push(arr);
    }
    return fn(n, i, j, res);
};

4. Student Attendance Record II LeetCode Solution Python

class Solution(object):
    def checkRecord(self, n):
        MOD = 10**9 + 7
        ax = x = xl = 1
        axl = axll = xll = 0
        for _ in range(n-1):
            ax, axl, axll, x, xl, xll =  (ax + x + xl + axl + axll + xll) %  MOD, ax, axl, (x + xl + xll) %  MOD, x, xl
        return (ax + x + xl + axl + axll + xll) %  MOD