LeetCode-in-Java
1191. K-Concatenation Maximum Sum
Medium
Given an integer array arr and an integer k, modify the array by repeating it k times.
For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].
Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0.
As the answer can be very large, return the answer modulo 109 + 7.
Example 1:
Input: arr = [1,2], k = 3
Output: 9
Example 2:
Input: arr = [1,-2,1], k = 5
Output: 2
Example 3:
Input: arr = [-1,-2], k = 7
Output: 0
Constraints:
1 <= arr.length <= 1051 <= k <= 105-104 <= arr[i] <= 104
Solution
public class Solution {
private long mod = 1000000007;
public int kConcatenationMaxSum(int[] arr, int k) {
// int kadane = Kadane(arr);
// #1 when k 1 simply calculate kadanes
if (k == 1) {
return (int) (kadane(arr) % mod);
}
// #2 else calculate the total sum and then check if sum is -Ve or +Ve
long totalSum = 0;
for (int i : arr) {
totalSum += i;
}
// #3 when negative then calculate of arr 2 times only the answer is in there only
if (totalSum < 0) {
// when -ve sum put a extra check here of max from 0
return (int) Math.max(kadaneTwo(arr) % mod, 0);
} else {
// #4 when sum is positve then the ans is kadane of 2 + sum * (k-2);
// these two are used sUm*(k-2) ensures that all other are also included
return (int) ((kadaneTwo(arr) + ((k - 2) * totalSum) + mod) % mod);
}
}
private long kadane(int[] arr) {
long max = arr[0];
long cur = 0;
for (int n : arr) {
cur += n;
max = Math.max(max, cur);
if (cur < 0) {
cur = 0;
}
}
return max;
}
private long kadaneTwo(int[] arr) {
long max = arr[0];
long cur = 0;
for (int i = 0; i < arr.length * 2; i++) {
cur += arr[i % arr.length];
max = Math.max(max, cur);
if (cur < 0) {
cur = 0;
}
}
return max;
}
}