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Maximum Product Subarray
The maximum product of sub-arrays in $[1, n]$ can be divided by 3 cases:
- A[n] is the maximum product of all sub-arrays in [1, n].
- The array which has the maximum product is end by A[n].
- The array of maximum product is not including A[n].
Thus the result can be expressed as
result = max(case1, case2, case3)
The second situation is not normal. If A[n] is positive, it can be got by prePositiveMax[n-1], if A[n] is negative, it can be got by preNegativeMin[n-1].
So one brute method is:
prePositiveMax[n] = max(prePositiveMax[n-1]*A[n], preNegativeMin[n-1]*A[n], A[n]); preNegativeMin[n] = min(prePositiveMax[n-1]*A[n], preNegativeMin[n-1]*A[n], A[n]);
A more accurate method is:
We define pEnd[i]: the maximum non-negative product of subarray with A[i]
We define nEnd[i]: the minimum non-positive product of subarray with A[i]
In fact, here we use pEnd, nEnd intead of prePositiveMax, preNegativeMin, thus we have
if(A[i] > 0){ pEnd[i] = max(A[i], pEnd[i-1]*A[i]); nEnd[i] = nEnd[i] * A[i]; }else{ pEnd[i] = nEnd[i] * A[i]; nEnd[i] = min(A[i], pEnd[i]*A[i]); }
Then, we can simplify as
if(A[i] < 0) swap(pEnd, nEnd);pEnd = max(pEnd*A[i], A[i]); nEnd = min(nEnd*A[i], A[i]);
So we conclude the whole code
int maxProduct(int A[], int n) { if(1 == n) return A[0]; int pEnd, nEnd, res; pEnd = nEnd = res = 0; for(int i = 0; i < n; ++i){ if(A[i] < 0) swap(&pEnd, &nEnd); pEnd = max(pEnd*A[i], A[i]); nEnd = min(nEnd*A[i], A[i]); if(res < pEnd) res = pEnd; //from res = max(res, pEnd) } return res;}
The complete code is:
class Solution {private: int max(int a, int b){ return a > b ? a : b; } int min(int a, int b){ return a > b ? b : a; } void swap(int* a, int* b){ *a = *a + *b; *b = *a - *b; *a = *a - *b; } public: int maxProduct(int A[], int n) { if(1 == n) return A[0]; int pEnd, nEnd, res; pEnd = nEnd = res = 0; for(int i = 0; i < n; ++i){ if(A[i] < 0) swap(&pEnd, &nEnd); pEnd = max(pEnd*A[i], A[i]); nEnd = min(nEnd*A[i], A[i]); if(res < pEnd) res = pEnd; } return res; }};
Reference
- https://oj.leetcode.com/discuss/11852/share-c-code-with-dp-o-1-space-o-n-time
- http://www.ahathinking.com/archives/120.html
Maximum Product Subarray
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