Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions:
- 0 < i, i + 1 < j, j + 1 < k < n - 1
- Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be equal.
where we define that subarray (L, R) represents a slice of the original array starting from the element indexed L to the element indexed R.
Example:
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| Input: [1,2,1,2,1,2,1]
Output: True
Explanation:
i = 1, j = 3, k = 5.
sum(0, i - 1) = sum(0, 0) = 1
sum(i + 1, j - 1) = sum(2, 2) = 1
sum(j + 1, k - 1) = sum(4, 4) = 1
sum(k + 1, n - 1) = sum(6, 6) = 1
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Note:
- 1 <= n <= 2000.
- Elements in the given array will be in range [-1,000,000, 1,000,000].
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| public class Solution {
public boolean splitArray(int[] nums) {
if (nums == null || nums.length < 7) {
return false;
}
int[] sum = new int[nums.length];
sum[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
sum[i] += sum[i - 1] + nums[i];
}
for (int j = 3; j < nums.length - 3; j++) {
Set<Integer> set = new HashSet<>();
for (int i = 1; i < j - 1; i++) {
if (sum[i - 1] == sum[j - 1] - sum[i]) {
set.add(sum[i - 1]);
}
}
for (int k = j + 1; k < nums.length - 1; k++) {
if (sum[k - 1] - sum[j] == sum[nums.length - 1] - sum[k] && set.contains(sum[k - 1] - sum[j])) {
return true;
}
}
}
return false;
}
}
|