LeetCode 805: Split Array With Same Average Solution
Master LeetCode problem 805 (Split Array With Same Average), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
805. Split Array With Same Average
Problem Explanation
Explanation
To solve this problem, we can use a dynamic programming approach. We need to check if it is possible to split the input array into two subarrays with equal average. We can iterate through the input array and calculate the sum of all elements. Then, we iterate through the possible lengths of the first subarray and check if we can achieve the same average in both subarrays by choosing elements for the first subarray.
- Create a 2D boolean array
dpto store whether it is possible to achieve a certain sum with a certain length. - Initialize
dp[0][0]as true, as it is always possible to achieve sum 0 with length 0. - Iterate through the input array and update the
dparray. - Check if it is possible to achieve the same average in both subarrays by checking if there exists a length
lensuch thatsum * len == totalSum * len1.
Time complexity: O(n * sum), where n is the number of elements in the input array and sum is the sum of all elements. Space complexity: O(n * sum).
Solution Code
class Solution {
public boolean splitArraySameAverage(int[] nums) {
int n = nums.length;
int sum = 0;
for (int num : nums) {
sum += num;
}
boolean[][] dp = new boolean[sum + 1][n / 2 + 1];
dp[0][0] = true;
for (int num : nums) {
for (int s = sum; s >= num; s--) {
for (int len = 1; len <= n / 2; len++) {
dp[s][len] |= dp[s - num][len - 1];
}
}
}
for (int len = 1; len <= n / 2; len++) {
if (sum * len % n == 0 && dp[sum * len / n][len]) {
return true;
}
}
return false;
}
}Try It Yourself
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