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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

• Java
• C++
• Python
• Go
• Javascript

• 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};
      }
  };
  

• 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]
  
  

• 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];
  };
  
  

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