https://h1ros.github.io/posts/coding/1031-maximum-sum-of-two-non-overlapping-subarrays/
X. Prefix sum
https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/discuss/278251/JavaC%2B%2BPython-O(N)Time-O(1)-Space
Lsum, sum of the last L elementsMsum, sum of the last M elementsLmax, max sum of contiguous L elements before the last M elements.Mmax, max sum of contiguous M elements before the last L elements/Complexity
Two pass,
O(N) time,O(1) extra space.
It can be done in one pass. I just don't feel like merging them.
If you don't like change the original input, don't have to.
If you don't like change the original input, don't have to.
Java:
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
for (int i = 1; i < A.length; ++i)
A[i] += A[i - 1];
int res = A[L + M - 1], Lmax = A[L - 1], Mmax = A[M - 1];
for (int i = L + M; i < A.length; ++i) {
Lmax = Math.max(Lmax, A[i - M] - A[i - L - M]);
Mmax = Math.max(Mmax, A[i - L] - A[i - L - M]);
res = Math.max(res, Math.max(Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L]));
}
return res;
}
Another Java
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
int res = 0, Lsum = 0, Lmax = 0, Msum = 0, Mmax = 0;
for (int i = 0; i < A.length; ++i) {
Msum += A[i];
if (i - M >= 0) Msum -= A[i - M];
if (i - M >= 0) Lsum += A[i - M];
if (i - M - L >= 0) Lsum -= A[i - L - M];
Lmax = Math.max(Lmax, Lsum);
res = Math.max(res, Lmax + Msum);
}
Lsum = Lmax = Msum = Mmax = 0;
for (int i = 0; i < A.length; ++i) {
Lsum += A[i];
if (i - L >= 0) Lsum -= A[i - L];
if (i - L >= 0) Msum += A[i - L];
if (i - M - L >= 0) Msum -= A[i - L - M];
Mmax = Math.max(Mmax, Msum);
res = Math.max(res, Mmax + Lsum);
}
return res;
}
}(前缀后缀预处理)
- 预处理前缀和数组
presum，后缀和数组sufsum。 - 用以上两个数组，预处理出前缀最大的
L子数组的和preL，前缀最大的M子数组的和preM，后缀最大的L子数组的和sufL，后缀最大的M子数组的和sufM。 - 根据两种情况
L前M后或者M前L后分别统计最大值。
X. https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/discuss/279221/Java-two-11-lines-easy-DP-codes-w-comment-time-O(n)-NO-change-of-input
- Scan the prefix sum array from index
L + M, which is the first possible position; - update the max value of the
L-lengthsubarray; then update max value of the sum of the both; - we need to swap
LandMto scan twice, since either subarray can occur before the other. - In private method, prefix sum difference
p[i - M] - p[i - M - L]isL-lengthsubarray from indexi - M - Ltoi - M - 1, andp[i] - p[i - M]isM-lengthsubarray from indexi - Mtoi - 1.
Solution 1:
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
int[] prefixSum = new int[A.length + 1];
for (int i = 0; i < A.length; ++i) {
prefixSum[i + 1] = prefixSum[i] + A[i];
}
return Math.max(maxSum(prefixSum, L, M), maxSum(prefixSum, M, L));
}
private int maxSum(int[] p, int L, int M) {
int ans = 0;
for (int i = L + M, maxL = 0; i < p.length; ++i) {
maxL = Math.max(maxL, p[i - M] - p[i - M - L]); // update max of L-length subarray.
ans = Math.max(ans, maxL + p[i] - p[i - M]); // update max of the sum of L-length & M-length subarrays.
}
return ans;
}
Analysis:
Time & space:
O(n), where n = A.length.
Solution 2:
Based on Solution 1, we can further get rid of prefix sum array to develop the following
O(1) code.
Unfortunately, the boundary conditions are headache and fallible, please let me know if you can improve the readability, or at least make it concise.
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
return Math.max(maxSum(A, L, M), maxSum(A, M, L));
}
private int maxSum(int[] A, int L, int M) {
int ans = 0;
for (int i = 0, maxL = 0, sumL = 0, sumM = 0; i < A.length; ++i) {
if (i < L || i >= L + M) { sumL += i < L ? A[i] : A[i - M]; } // first L-length subarray at index [0...L - 1], no update between index [L...L + M - 1].
if (i >= L) { sumM += A[i]; } // first M-length subarray starts from index L to L + M - 1.
if (i >= L + M) { sumL -= A[i - L - M]; } // deduct first item from current L-length subarray.
if (i >= L + M) { sumM -= A[i - M]; } // deduct first item from current M-length subarray.
if (i >= L + M - 1) { maxL = Math.max(maxL, sumL); } // update max of L-length subarray.
if (i >= L + M - 1) { ans = Math.max(ans, maxL + sumM); } // update max of L-length & M-length subarrays.
}
return ans;
}
https://leetcode.com/problems/maximum-sum-of-two-non-overlapping-subarrays/discuss/288993/Java-DP-solution
Extend Kadane’s algorithm for two non-overlapping subarrays with L & M in length
/* |---L--| <- prior maxL |---L--| <- sum[i-M]-sum[i-M-L] |---M----| --------------------|--------i |--M---| <- maxM at i-L position |---L----| --------------------|--------i */ public int maxSumTwoNoOverlap(int[] A, int L, int M) { if (A==null || A.length < L+M) return 0; int []sum = new int[A.length]; sum[0]=A[0]; for(int i=1; i< A.length;i++){ sum[i] = sum[i-1] + A[i]; } int maxL = sum[L-1]; // initial value int maxM = sum[M-1]; // initial value int res = sum[L+M-1]; // initial value for(int i=L+M; i < sum.length; i++){ maxL = Math.max(maxL, sum[i-M]-sum[i-M-L]); // maxL at i-M position maxM = Math.max(maxM, sum[i-L]-sum[i-M-L]); //maxM at i-L position res = Math.max(res, Math.max(maxL + sum[i] - sum[i-M], maxM + sum[i] - sum[i-L])); } return res; }
https://www.acwing.com/solution/LeetCode/content/1814/
给出非负整数数组
A，返回两个非重叠（连续）子数组中元素的最大和，子数组的长度分别为 L 和 M。（这里需要澄清的是，长为 L 的子数组可以出现在长为 M 的子数组之前或之后。）
正式地，返回最大的
V，而 V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) 并满足下列条件之一：0 <= i < i + L - 1 < j < j + M - 1 < A.length，或0 <= j < j + M - 1 < i < i + L - 1 < A.length。
样例
输入：A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
输出：20
解释：子数组的一种选择中，[9] 长度为 1，[6,5] 长度为 2。
输入：A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
输出：29
解释：子数组的一种选择中，[3,8,1] 长度为 3，[8,9] 长度为 2。
输入：A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
输出：31
解释：子数组的一种选择中，[5,6,0,9] 长度为 4，[0,3,8] 长度为 3。
X. 花花酱 LeetCode Weekly Contest 133 (1029,1030,1031,1032)
