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package g1101_1200.s1191_k_concatenation_maximum_sum;
// #Medium #Array #Dynamic_Programming #2022_03_03_Time_6_ms_(73.85%)_Space_59.8_MB_(30.38%)
public class Solution {
private static final 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;
}
}
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