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
1 | 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:
1 2 3 | 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
1 2 3 4 5 6 7 | 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
1 2 3 4 | 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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 | 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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 | 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; } }; |
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