LeetCode-in-Java
805. Split Array With Same Average
Hard
You are given an integer array nums.
You should move each element of nums into one of the two arrays A and B such that A and B are non-empty, and average(A) == average(B).
Return true if it is possible to achieve that and false otherwise.
Note that for an array arr, average(arr) is the sum of all the elements of arr over the length of arr.
Example 1:
Input: nums = [1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have an average of 4.5.
Example 2:
Input: nums = [3,1]
Output: false
Constraints:
1 <= nums.length <= 300 <= nums[i] <= 104
Solution
import java.util.Arrays;
public class Solution {
private int[] nums;
private int[] sums;
public boolean splitArraySameAverage(int[] nums) {
int len = nums.length;
if (len == 1) {
return false;
}
Arrays.sort(nums);
sums = new int[len + 1];
for (int i = 0; i < len; i++) {
sums[i + 1] = sums[i] + nums[i];
}
int sum = sums[len];
this.nums = nums;
for (int i = 1, stop = len / 2; i <= stop; i++) {
if ((sum * i) % len == 0 && findSum(i, len, (sum * i) / len)) {
return true;
}
}
return false;
}
private boolean findSum(int k, int pos, int target) {
if (k == 1) {
while (true) {
if (nums[--pos] <= target) {
break;
}
}
return nums[pos] == target;
}
for (int i = pos; sums[i] - sums[i-- - k] >= target; ) {
if (sums[k - 1] <= target - nums[i] && findSum(k - 1, i, target - nums[i])) {
return true;
}
}
return false;
}
}