Leetcode 1588 sum of all ODD length subarrays. C++ - c++

Question

I'm practicing myself by doing some leetcode questions, however, I don't know why that's an overflow problem right here. I knew the way I sum the subarray was terrible, any tips for the sum of the subarray?
and the run time for this code would be forever
#include <numeric>
class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int size = arr.size();//5
int ans = 0;
int sumAll = 0;
int start = 3;
int tempsum;
for(int i =0; i< size; i++){ //sumitself
sumAll += arr[i];
}
ans = sumAll; //alreayd have the 1 index
if(size%2 == 0){//even number 6
int temp = size-1; //5
if(size == 2)
ans = sumAll;
else{
while(start <= temp){//3 < 5
for(int i = 0; i< size; i++){
for(int k =0; k< start; k++){//3
tempsum += arr[i+k];
if(i+k > temp) //reach 5
break;
}
}
start+=2;
}
}
ans+= tempsum;
}
else{//odd number
if(size == 1)
ans = sumAll;
else{
while(start < size){//3
for(int i = 0; i< size; i++){
for(int k =0; k< start; k++){//3
tempsum += arr[i+k];
if(i+k > size) //reach 5
break;
}
}
start+=2;
}
ans+= tempsum;
ans+= sumAll; //size index
}
}
return ans;
}
};

Answer 1

The problem is with arr[i+k]. The result of i + k can be equal to, or larger, than size. You check it after you have already gone out of bounds.
You should probably modify the inner loop condition so that never happens:
for(int k =0; k < start && (i + k) < size; k++){//3
Now you don't even need the inner check.

Answer 2

You can use prefix sum array technique and then for each index you can calculate the sub-array sum for each odd-length array using prefix sum array. I submitted the below solution in LeetCode and it beats runtime of 100% of submissions and memory usage of 56.95%
class Solution {
public:
int sumOddLengthSubarrays(vector<int>& arr) {
int n = arr.size();
vector<int> prefix(n+1,0);
int sum = 0;
prefix[1] = arr[0];
for(int i=1;i<n;i++)
prefix[i+1]=(arr[i]+prefix[i]);
for(int i=0;i<n;i++)
{
for(int j=i;j<n;j+=2)
sum+=prefix[j+1]-prefix[i];
}
return sum;
}
};

Answer 3

https://leetcode.com/problems/sum-of-all-odd-length-subarrays/discuss/1263893/Java-100-one-pass-O(n)-with-explanation
class Solution {
public int sumOddLengthSubarrays(int[] arr) {
// alt solution: O(n)
//for each i:
// if(n -1 - i) is odd, then arr[i] is counted (n-1-i)/2 + 1 times, each from 0 to i, total ((n-i)/2+1)*(i+1) times
// if(n -1 - i) is even, then arr[i] is counted (n-1-i)/2 + 1 times, if starting subseq index diff with i is even;
// (n-1-i)/2 times, if starting index diff with i s odd, total (n-i)/2 *(i+1) + (i+1)/2
// if i is even i - 1, i - 3, .. 1, total (i -2)/2 + 1 = i / 2 = (i+1) / 2
// if i is odd i-1, i-3, .., 0 total (i-1)/2 + 1 = (i+1) / 2
int total = 0;
int n = arr.length;
for(int i = 0; i < n; i++)
total += (((n - 1 - i) / 2 + 1) * (i + 1) - ((n-i) % 2)*((i+1) / 2)) * arr[i];
return total;
}
}