links nach rechts.
Code: Select all
Array = [2,3,2,1]
< /code>
Die Sub -Arrays sind: [start_index, end_index] < /p>
[0,0] subarray = [2], min is 2 and total of items = 2. min * total = 2*2=4
[0,1] subarray = [2,3], min is 2 and total of items = 5. min * total = 2*5=10
[0,2] subarray = [2,3,2], min is 2 and total of items = 7. min * total = 2*7=14
[0,3] subarray = [2,3,2,1], min is 1 and total of items = 8. min * total = 1*8=8
[1,1] subarray = [3], min is 3 and total of items = 3. min * total = 3*3 = 9
[1,2] subarray = [3,2], min is 2 and total of items = 5. min * total = 2*5 = 10
[1,3] subarray = [3,2,1], min is 1 and total of items = 6. min * total = 1*6 = 6
[2,2] subarray = [2], min is 2 and total of items = 2. min * total = 2*2 = 4
[2,3] subarray = [2,1], min is 1 and total of items = 3. min * total = 1*3 = 3
[3,3] subarray = [1], min is 1 and total of items = 1. min * total = 1*1 = 1
Total = 4 + 10 + 14 + 8 + 9 + 10+ 6 + 4 + 3 + 1 = 69
Code: Select all
Each array element is in range 1 to 10^9. Array size 1 to 10^5. Return response in modulo 10^9+7
< /code>
Dies ist der Code, den ich ausprobiert habe. < /p>
public static int process(List list) {
int n = list.size();
int mod = 7 + 1000_000_000;
long result = 0;
for (int i = 0; i < n; i++) {
long total = 0;
int min = list.get(i);
for (int j = i; j < n; j++) {
int p = list.get(j);
total = (total + p) % mod;
min = Math.min(min, p);
result = (result + (min * total) % mod) % mod;
}
}
return (int) result;
}
Was kann ein besserer Ansatz sein, um diese Aufgabe zu lösen?
Code: Select all
David Eisenstat
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