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| 1 | +/** |
| 2 | + * [1573] Number of Ways to Split a String |
| 3 | + * |
| 4 | + * Given a binary string s, you can split s into 3 non-empty strings s1, s2, and s3 where s1 + s2 + s3 = s. |
| 5 | + * Return the number of ways s can be split such that the number of ones is the same in s1, s2, and s3. Since the answer may be too large, return it modulo 10^9 + 7. |
| 6 | + * |
| 7 | + * Example 1: |
| 8 | + * |
| 9 | + * Input: s = "10101" |
| 10 | + * Output: 4 |
| 11 | + * Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'. |
| 12 | + * "1|010|1" |
| 13 | + * "1|01|01" |
| 14 | + * "10|10|1" |
| 15 | + * "10|1|01" |
| 16 | + * |
| 17 | + * Example 2: |
| 18 | + * |
| 19 | + * Input: s = "1001" |
| 20 | + * Output: 0 |
| 21 | + * |
| 22 | + * Example 3: |
| 23 | + * |
| 24 | + * Input: s = "0000" |
| 25 | + * Output: 3 |
| 26 | + * Explanation: There are three ways to split s in 3 parts. |
| 27 | + * "0|0|00" |
| 28 | + * "0|00|0" |
| 29 | + * "00|0|0" |
| 30 | + * |
| 31 | + * |
| 32 | + * Constraints: |
| 33 | + * |
| 34 | + * 3 <= s.length <= 10^5 |
| 35 | + * s[i] is either '0' or '1'. |
| 36 | + * |
| 37 | + */ |
| 38 | +pub struct Solution {} |
| 39 | + |
| 40 | +// problem: https://leetcode.com/problems/number-of-ways-to-split-a-string/ |
| 41 | +// discuss: https://leetcode.com/problems/number-of-ways-to-split-a-string/discuss/?currentPage=1&orderBy=most_votes&query= |
| 42 | + |
| 43 | +// submission codes start here |
| 44 | + |
| 45 | +impl Solution { |
| 46 | + // Credit: https://leetcode.com/problems/number-of-ways-to-split-a-string/solutions/3174473/just-a-runnable-solution/ |
| 47 | + |
| 48 | + const MOD: i32 = 1_000_000_007; |
| 49 | + |
| 50 | + pub fn num_ways(s: String) -> i32 { |
| 51 | + let s = s.as_bytes(); |
| 52 | + let n = s.len(); |
| 53 | + let mut one = 0; |
| 54 | + let result; |
| 55 | + |
| 56 | + for &item in s.iter().take(n) { |
| 57 | + if item == b'1' { |
| 58 | + one += 1; |
| 59 | + } |
| 60 | + } |
| 61 | + |
| 62 | + if one % 3 != 0 { |
| 63 | + return 0; |
| 64 | + } else if one == 0 { |
| 65 | + result = ((n as i64 - 1) * (n as i64 - 2) / 2) % MOD; |
| 66 | + } else { |
| 67 | + let count_of_one = one / 3; |
| 68 | + let (mut index1, mut index1end, mut index2, mut index2end) = (-1, -1, -1, -1); |
| 69 | + let mut count = 0; |
| 70 | + for (i, &item) in s.iter().enumerate().take(n) { |
| 71 | + if item == b'1' { |
| 72 | + count += 1; |
| 73 | + } |
| 74 | + if count == count_of_one && index1 == -1 { |
| 75 | + index1 = i as i64; |
| 76 | + } |
| 77 | + if count == count_of_one + 1 && index2 == -1 { |
| 78 | + index2 = i as i64; |
| 79 | + } |
| 80 | + if count == count_of_one * 2 && index1end == -1 { |
| 81 | + index1end = i as i64; |
| 82 | + } |
| 83 | + if count == count_of_one * 2 + 1 && index2end == -1 { |
| 84 | + index2end = i as i64; |
| 85 | + } |
| 86 | + } |
| 87 | + result = ((index2 - index1) % MOD) * ((index2end - index1end) % MOD) % MOD; |
| 88 | + } |
| 89 | + |
| 90 | + result as _ |
| 91 | + } |
| 92 | +} |
| 93 | + |
| 94 | +// submission codes end |
| 95 | + |
| 96 | +#[cfg(test)] |
| 97 | +mod tests { |
| 98 | + use super::*; |
| 99 | + |
| 100 | + #[test] |
| 101 | + fn test_1573_example_1() { |
| 102 | + let s = "10101".to_string(); |
| 103 | + |
| 104 | + let result = 4; |
| 105 | + |
| 106 | + assert_eq!(Solution::num_ways(s), result); |
| 107 | + } |
| 108 | + |
| 109 | + #[test] |
| 110 | + fn test_1573_example_2() { |
| 111 | + let s = "1001".to_string(); |
| 112 | + |
| 113 | + let result = 0; |
| 114 | + |
| 115 | + assert_eq!(Solution::num_ways(s), result); |
| 116 | + } |
| 117 | + |
| 118 | + #[test] |
| 119 | + fn test_1573_example_3() { |
| 120 | + let s = "0000".to_string(); |
| 121 | + |
| 122 | + let result = 3; |
| 123 | + |
| 124 | + assert_eq!(Solution::num_ways(s), result); |
| 125 | + } |
| 126 | +} |
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