LeetCode-in-Java
2007. Find Original Array From Doubled Array
Medium
An integer array original is transformed into a doubled array changed by appending twice the value of every element in original, and then randomly shuffling the resulting array.
Given an array changed, return original if changed is a doubled array. If changed is not a doubled array, return an empty array. The elements in original may be returned in any order.
Example 1:
Input: changed = [1,3,4,2,6,8]
Output: [1,3,4]
Explanation: One possible original array could be [1,3,4]:
-
Twice the value of 1 is 1 * 2 = 2.
-
Twice the value of 3 is 3 * 2 = 6.
-
Twice the value of 4 is 4 * 2 = 8.
Other original arrays could be [4,3,1] or [3,1,4].
Example 2:
Input: changed = [6,3,0,1]
Output: []
Explanation: changed is not a doubled array.
Example 3:
Input: changed = [1]
Output: []
Explanation: changed is not a doubled array.
Constraints:
1 <= changed.length <= 1050 <= changed[i] <= 105
Solution
public class Solution {
public int[] findOriginalArray(int[] changed) {
if (changed.length % 2 == 1) {
return new int[0];
}
int[] a = new int[100001];
for (int j : changed) {
a[j]++;
}
if (a[0] % 2 == 1) {
return new int[0];
}
int[] ans = new int[changed.length / 2];
int p = 0;
if (a[0] > 0) {
a[0] /= 2;
while (a[0] > 0) {
ans[p++] = 0;
a[0]--;
}
}
for (int i = 1; i <= 100001 / 2; i++) {
if (a[i] == 0) {
continue;
}
int tmp = i * 2;
if (a[tmp] >= a[i]) {
a[tmp] = a[tmp] - a[i];
while (a[i] > 0) {
ans[p++] = i;
a[i]--;
}
} else {
return new int[0];
}
}
for (int i = 1; i < a.length; i++) {
if (a[i] != 0) {
return new int[0];
}
}
return ans;
}
}