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maximum_subarray.cpp
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124 lines (101 loc) · 3 KB
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/*
Maximum Subarray
Find the contiguous subarray within an array (containing at least one number)
which has the largest sum.
For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.
More practice:
If you have figured out the O(n) solution, try coding another solution
using the divide and conquer approach, which is more subtle.
*/
#include <iostream>
#include <climits>
#include <initializer_list>
#include <algorithm>
using namespace std;
class Solution
{
public:
int maxSubArray(int A[], int n)
{
int max_sum = INT_MIN;
int interimSum = INT_MIN; // interim sum ending with array element i
// dp
for (int i = 0; i < n; i++) {
if (interimSum <= 0)
interimSum = A[i];
else
interimSum += A[i];
// max sum is the max of all interimSums
max_sum = max(interimSum, max_sum);
}
return max_sum;
}
int maxSubArray2(int A[], int n)
{
int max_sum = INT_MIN;
int sum = 0;
for (int i = 0; i < n; i++) {
sum = max(A[i], sum + A[i]);
max_sum = max(sum, max_sum);
}
return max_sum;
}
};
// divide and conquer
// three cases:
// 1) max is on the left
// 2) max is on the right
// 3) max includes middle element
class Solution2
{
public:
int divideAndConquer(int A[], int l, int h)
{
if (l > h)
return 0;
else if (l == h)
return A[l];
else {
int mid = l + ((h - l) >> 1);
int leftSum = 0;
int leftMax = INT_MIN;
for (int k = mid; k >= l; k--) {
leftSum += A[k];
leftMax = max(leftMax, leftSum);
}
int rightSum = 0;
int rightMax = INT_MIN;
for (int k = mid + 1; k <= h; k++) {
rightSum += A[k];
rightMax = max(rightMax, rightSum);
}
int lMax = divideAndConquer(A, l, mid);
int rMax = divideAndConquer(A, mid + 1, h);
return max( { lMax, rMax, leftMax + rightMax });
}
}
int maxSubArray(int A[], int n)
{
if (n == 0)
return 0;
else {
return divideAndConquer(A, 0, n - 1);
}
}
};
int main(int argc, char *argv[])
{
int array[] = { -1, 1, -3, 4, -1, 2, 1, -5, 4 };
int array2[] = { -1, 0, -2 };
int array3[] = { 1, 2, -1 };
Solution sol;
cout << sol.maxSubArray(array, sizeof(array) / sizeof(int)) << endl;
cout << sol.maxSubArray(array2, sizeof(array2) / sizeof(int)) << endl;
cout << sol.maxSubArray(array3, sizeof(array3) / sizeof(int)) << endl;
Solution2 sol2;
cout << sol2.maxSubArray(array, sizeof(array) / sizeof(int)) << endl;
cout << sol2.maxSubArray(array2, sizeof(array2) / sizeof(int)) << endl;
cout << sol2.maxSubArray(array3, sizeof(array3) / sizeof(int)) << endl;
return 0;
}