min cost of climbing stairs dynamic programming - c++

Question

I am trying to solve the following problem on leetcode:
On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).
Once you pay the cost, you can either climb one or two steps.
You need to find minimum cost to reach the top of the floor, and you can either start
from the step with index 0, or the step with index 1.
My solution is shown below:
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
return helper(cost, cost.size() - 1);
}
int helper(vector<int>& cost, int currStair) {
static vector<double> minCost(cost.size(), 0);
minCost[0] = cost[0];
minCost[1] = cost[1];
if (minCost[currStair] > 0) {
return minCost[currStair];
}
return minCost[currStair] = min(helper(cost, currStair - 1), helper(cost, currStair - 2)) + cost[currStair];
}
};
When I try to submit, I get the following run-time error. Why?
AddressSanitizer: heap-buffer-overflow on address 0x603000000008 at pc 0x0000004089af bp 0x7ffdd02dcaa0 sp 0x7ffdd02dca98
EDITED SOLUTION:
class Solution {
public:
int minCostClimbingStairs(vector<int>& cost) {
return helper(cost, cost.size() - 1);
}
int helper(vector<int>& cost, int currStair) {
if (currStair < 0 || cost.size() <= 1) {
return 0;
}
static vector<double> minCost(cost.size(), 0);
minCost[0] = cost[0];
minCost[1] = cost[1];
if (minCost[currStair] > 0) {
return minCost[currStair];
}
minCost[currStair] = min(helper(cost, currStair - 1), helper(cost, currStair - 2)) + cost[currStair];
return min(minCost[cost.size()-1], minCost[cost.size()-2]);
}
};
As you can see I made changes in the end of the code

Answer 1

The problem is a missing base case for the recursion. currStair keeps decrementing by -1 or -2 in each recursive call, but there's no condition to check if it went below 0 and cut off the recursion, resulting in an illegal memory access like minCost[-1].
Add the precondition
if (currStair < 0) return 0;
and you're back on track, although the algorithm still has correctness issues to resolve.
Here's a hint that might help you get unstuck:
The instructions say: "You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1."