927. Three Equal Parts

Description

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 * 10⁴
arr[i] is 0 or 1

Solutions

Solution 1: Counting + Three Pointers

We denote the length of the array as $n$, and the number of '1's in the array as $cnt$.

Obviously, $cnt$ must be a multiple of $3$, otherwise the array cannot be divided into three equal parts, and we can return $[-1, -1]$ in advance. If $cnt$ is $0$, it means that all elements in the array are '0', and we can directly return $[0, n - 1]$.

We divide $cnt$ by $3$ to get the number of '1's in each part, and then find the position of the first '1' in each part in the array arr (refer to the $find(x)$ function in the following code), denoted as $i$, $j$, $k$ respectively.

0 1 1 0 0 0 1 1 0 0 0 0 0 1 1 0 0
  ^         ^             ^
  i         j             k

Then we start from $i$, $j$, $k$ and traverse each part at the same time, check whether the corresponding values of the three parts are equal. If they are, continue to traverse until $k$ reaches the end of arr.

0 1 1 0 0 0 1 1 0 0 0 0 0 1 1 0 0
          ^         ^             ^
          i         j             k

At the end of the traversal, if $k=n$, it means that it satisfies the three equal parts, and we return $[i - 1, j]$ as the answer, otherwise return $[-1, -1]$.

The time complexity is $O(n)$, where $n$ is the length of arr. The space complexity is $O(1)$.

Similar problems:

1573. Number of Ways to Split a String

Python3JavaC++GoJavaScript

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]

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;
    }
}

class Solution {
public:
    vector<int> threeEqualParts(vector<int>& arr) {
        int n = arr.size();
        int cnt = accumulate(arr.begin(), arr.end(), 0);
        if (cnt % 3) return {-1, -1};
        if (!cnt) return {0, n - 1};
        cnt /= 3;

        auto find = [&](int x) {
            int s = 0;
            for (int i = 0; i < n; ++i) {
                s += arr[i];
                if (s == x) return i;
            }
            return 0;
        };
        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 ? vector<int>{i - 1, j} : vector<int>{-1, -1};
    }
};

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}
}

/**
 * @param {number[]} arr
 * @return {number[]}
 */
var threeEqualParts = function (arr) {
    function find(x) {
        let s = 0;
        for (let i = 0; i < n; ++i) {
            s += arr[i];
            if (s == x) {
                return i;
            }
        }
        return 0;
    }
    const n = arr.length;
    let cnt = 0;
    for (const v of arr) {
        cnt += v;
    }
    if (cnt % 3) {
        return [-1, -1];
    }
    if (cnt == 0) {
        return [0, n - 1];
    }
    cnt = Math.floor(cnt / 3);
    let [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) {}
    return k == n ? [i - 1, j] : [-1, -1];
};

927. Three Equal Parts

Description

Solutions

Solution 1: Counting + Three Pointers

Comments