Three Equal Parts

// Accepted solution for LeetCode #927: Three Equal Parts
class Solution {
    private int[] arr;

    public int[] threeEqualParts(int[] arr) {
        this.arr = arr;
        int cnt = 0;
        int n = arr.length;
        for (int v : arr) {
            cnt += v;
        }
        if (cnt % 3 != 0) {
            return new int[] {-1, -1};
        }
        if (cnt == 0) {
            return new int[] {0, n - 1};
        }
        cnt /= 3;

        int i = find(1), j = find(cnt + 1), k = find(cnt * 2 + 1);
        for (; k < n && arr[i] == arr[j] && arr[j] == arr[k]; ++i, ++j, ++k) {
        }
        return k == n ? new int[] {i - 1, j} : new int[] {-1, -1};
    }

    private int find(int x) {
        int s = 0;
        for (int i = 0; i < arr.length; ++i) {
            s += arr[i];
            if (s == x) {
                return i;
            }
        }
        return 0;
    }
}

// Accepted solution for LeetCode #927: Three Equal Parts
func threeEqualParts(arr []int) []int {
	find := func(x int) int {
		s := 0
		for i, v := range arr {
			s += v
			if s == x {
				return i
			}
		}
		return 0
	}
	n := len(arr)
	cnt := 0
	for _, v := range arr {
		cnt += v
	}
	if cnt%3 != 0 {
		return []int{-1, -1}
	}
	if cnt == 0 {
		return []int{0, n - 1}
	}
	cnt /= 3
	i, j, k := find(1), find(cnt+1), find(cnt*2+1)
	for ; k < n && arr[i] == arr[j] && arr[j] == arr[k]; i, j, k = i+1, j+1, k+1 {
	}
	if k == n {
		return []int{i - 1, j}
	}
	return []int{-1, -1}
}

# Accepted solution for LeetCode #927: Three Equal Parts
class Solution:
    def threeEqualParts(self, arr: List[int]) -> List[int]:
        def find(x):
            s = 0
            for i, v in enumerate(arr):
                s += v
                if s == x:
                    return i

        n = len(arr)
        cnt, mod = divmod(sum(arr), 3)
        if mod:
            return [-1, -1]
        if cnt == 0:
            return [0, n - 1]

        i, j, k = find(1), find(cnt + 1), find(cnt * 2 + 1)
        while k < n and arr[i] == arr[j] == arr[k]:
            i, j, k = i + 1, j + 1, k + 1
        return [i - 1, j] if k == n else [-1, -1]

// Accepted solution for LeetCode #927: Three Equal Parts
/**
 * [0927] Three Equal Parts
 *
 * You are given an array arr which consists of only zeros and ones, divide the array into three non-empty parts such that all of these parts represent the same binary value.
 * If it is possible, return any [i, j] with i + 1 < j, such that:
 *
 * 	arr[0], arr[1], ..., arr[i] is the first part,
 * 	arr[i + 1], arr[i + 2], ..., arr[j - 1] is the second part, and
 * 	arr[j], arr[j + 1], ..., arr[arr.length - 1] is the third part.
 * 	All three parts have equal binary values.
 *
 * If it is not possible, return [-1, -1].
 * Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.
 *  
 * Example 1:
 * Input: arr = [1,0,1,0,1]
 * Output: [0,3]
 * Example 2:
 * Input: arr = [1,1,0,1,1]
 * Output: [-1,-1]
 * Example 3:
 * Input: arr = [1,1,0,0,1]
 * Output: [0,2]
 *  
 * Constraints:
 *
 * 	3 <= arr.length <= 3 * 10^4
 * 	arr[i] is 0 or 1
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/three-equal-parts/
// discuss: https://leetcode.com/problems/three-equal-parts/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/three-equal-parts/solutions/1343847/rust-solution/
    pub fn three_equal_parts(arr: Vec<i32>) -> Vec<i32> {
        let v = arr
            .iter()
            .enumerate()
            .filter_map(|(i, &a)| if a == 1 { Some(i) } else { None })
            .collect::<Vec<_>>();
        if arr.len() < 3 || v.len() % 3 != 0 {
            return [-1, -1].to_vec();
        }
        if v.is_empty() {
            return [0, 2].to_vec();
        }
        let chunks = v.chunks(v.len() / 3).collect::<Vec<_>>();
        let i = chunks[0][v.len() / 3 - 1] + arr.len() - v[v.len() - 1] - 1;
        let j = chunks[1][v.len() / 3 - 1] + arr.len() - v[v.len() - 1];
        if arr.len() - chunks[2][0] > j - i - 1 || arr.len() - chunks[2][0] > i + 1 {
            return [-1, -1].to_vec();
        }
        for k in 0..(i + 1).min(j - i - 1).min(arr.len() - j) {
            if arr[i - k] != arr[j - 1 - k] || arr[i - k] != arr[arr.len() - 1 - k] {
                return [-1, -1].to_vec();
            }
        }
        [i as i32, j as i32].to_vec()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_0927_example_1() {
        let arr = vec![1, 0, 1, 0, 1];
        let result = vec![0, 3];

        assert_eq!(Solution::three_equal_parts(arr), result);
    }

    #[test]
    fn test_0927_example_2() {
        let arr = vec![1, 1, 0, 1, 1];
        let result = vec![-1, -1];

        assert_eq!(Solution::three_equal_parts(arr), result);
    }

    #[test]
    fn test_0927_example_3() {
        let arr = vec![1, 1, 0, 0, 1];
        let result = vec![0, 2];

        assert_eq!(Solution::three_equal_parts(arr), result);
    }
}

// Accepted solution for LeetCode #927: Three Equal Parts
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #927: Three Equal Parts
// class Solution {
//     private int[] arr;
// 
//     public int[] threeEqualParts(int[] arr) {
//         this.arr = arr;
//         int cnt = 0;
//         int n = arr.length;
//         for (int v : arr) {
//             cnt += v;
//         }
//         if (cnt % 3 != 0) {
//             return new int[] {-1, -1};
//         }
//         if (cnt == 0) {
//             return new int[] {0, n - 1};
//         }
//         cnt /= 3;
// 
//         int i = find(1), j = find(cnt + 1), k = find(cnt * 2 + 1);
//         for (; k < n && arr[i] == arr[j] && arr[j] == arr[k]; ++i, ++j, ++k) {
//         }
//         return k == n ? new int[] {i - 1, j} : new int[] {-1, -1};
//     }
// 
//     private int find(int x) {
//         int s = 0;
//         for (int i = 0; i < arr.length; ++i) {
//             s += arr[i];
//             if (s == x) {
//                 return i;
//             }
//         }
//         return 0;
//     }
// }