LeetCode in Kotlin
1013. Partition Array Into Three Parts With Equal Sum
Easy
Given an array of integers arr, return true if we can partition the array into three non-empty parts with equal sums.
Formally, we can partition the array if we can find indexes i + 1 < j with (arr[0] + arr[1] + ... + arr[i] == arr[i + 1] + arr[i + 2] + ... + arr[j - 1] == arr[j] + arr[j + 1] + ... + arr[arr.length - 1])
Example 1:
Input: arr = [0,2,1,-6,6,-7,9,1,2,0,1]
Output: true
Explanation: 0 + 2 + 1 = -6 + 6 - 7 + 9 + 1 = 2 + 0 + 1
Example 2:
Input: arr = [0,2,1,-6,6,7,9,-1,2,0,1]
Output: false
Example 3:
Input: arr = [3,3,6,5,-2,2,5,1,-9,4]
Output: true
Explanation: 3 + 3 = 6 = 5 - 2 + 2 + 5 + 1 - 9 + 4
Constraints:
3 <= arr.length <= 5 * 104-104 <= arr[i] <= 104
Solution
class Solution {
fun canThreePartsEqualSum(arr: IntArray): Boolean {
var sum = 0
for (j in arr) {
sum += j
}
// 1. Base condition , the sum should be equally divided into 3 parts
if (sum % 3 != 0) {
return false
}
val eq = sum / 3
// to keep track of occurences of sum in the sub array
var count = 0
var temp = 0
for (j in arr) {
// 2. Base / Break condition for loop , i.e. if the count is 2,
// i.e. sum has been achieved twice ( and there is more indices
// to go through since we are in the loop again ) then return true
if (count == 2) {
return true
}
// 3. Adding to temp array
temp += j
// 4. If sum is achieved , increase the count
if (temp == eq) {
count++
// put temp=0 to start summing up from the next indices
temp = 0
}
}
// 5. If the above conditoin fails , result is false
return false
}
}