重叠的连续子数组的K个最大和
原文： https://www.geeksforgeeks.org/k-maximum-sum-overlapping-contiguous-sub-arrays/
给定一个整数数组和一个整数值k，找出k个最大和为k的子数组（可能重叠）。
示例：
Input : arr = {4, -8, 9, -4, 1, -8, -1, 6}, k = 4
Output : 9 6 6 5
Input : arr = {-2, -3, 4, -1, -2, 1, 5, -3}, k= 3
Output : 7 6 5
使用 Kadane 的算法，我们可以找到数组的最大连续子数组和。 但在一些情况下，Kadane 的算法不起作用。 当我们在数组中命中负数时，会将max_ending_here变量设置为零，因此我们错过了第二大值的可能性。
这里我们是由 Sung Eun Bae 和 Tadao Takaoka 提出的算法，该算法计算O(n)时间中的最大子数组和问题和O(k * n)时间中的k个最大子数组和问题。
首先，我们看一下使用这种方法的最大子数组和的问题：
先决条件：
k个最大子数组的方法：
1\. Calculate the prefix sum of the input array.
2\. Take cand, maxi and mini as arrays of size k.
3\. Initialize mini[0] = 0 for the same reason as previous.
4\. for each value of the prefix_sum[i] do
(i). update cand[j] value by prefix_sum[i] - mini[j]
(ii). maxi will be the maximum k elements of maxi and cand
(iii). if prefix_sum is minimum than all values of mini then
include it in mini and remove maximum element form mini
// After the ith iteration mini holds k minimum prefix sum upto
// index i and maxi holds k maximum overlapping sub-array sums
// upto index i.
5\. return maxi
在整个计算方法中，我们将maxi保持不变，而mini则保持不变。
C++
// C++ program to find out k maximum sum of
// overlapping sub-arrays
#include <iostream>
#include <limits>
#include <vector>
using namespace std;
// Function to compute prefix-sum of the input array
vector<int> prefix_sum(vector<int> arr, int n)
{
vector<int> pre_sum;
pre_sum.push_back(arr[0]);
for (int i = 1; i < n; i++)
pre_sum.push_back(pre_sum[i - 1] + arr[i]);
return pre_sum;
}
// Update maxi by k maximum values from maxi and cand
void maxMerge(vector<int>& maxi, vector<int> cand)
{
// Here cand and maxi arrays are in non-increasing
// order beforehand. Now, j is the index of the
// next cand element and i is the index of next
// maxi element. Traverse through maxi array.
// If cand[j] > maxi[i] insert cand[j] at the ith
// position in the maxi array and remove the minimum
// element of the maxi array i.e. the last element
// and increase j by 1 i.e. take the next element
// from cand.
int k = maxi.size();
int j = 0;
for (int i = 0; i < k; i++) {
if (cand[j] > maxi[i]) {
maxi.insert(maxi.begin() + i, cand[j]);
maxi.erase(maxi.begin() + k);
j += 1;
}
}
}
// Insert prefix_sum[i] to mini array if needed
void insertMini(vector<int>& mini, int pre_sum)
{
// Traverse the mini array from left to right.
// If prefix_sum[i] is less than any element
// then insert prefix_sum[i] at that position
// and delete maximum element of the mini array
// i.e. the rightmost element from the array.
int k = mini.size();
for (int i = 0; i < k; i++) {
if (pre_sum < mini[i]) {
mini.insert(mini.begin() + i, pre_sum);
mini.erase(mini.begin() + k);
break;
}
}
}
// Function to compute k maximum overlapping sub-
// array sums
void kMaxOvSubArray(vector<int> arr, int k)
{
int n = arr.size();
// Compute the prefix sum of the input array.
vector<int> pre_sum = prefix_sum(arr, n);
// Set all the elements of mini as +infinite
// except 0th. Set the 0th element as 0\.
vector<int> mini(k, numeric_limits<int>::max());
mini[0] = 0;
// Set all the elements of maxi as -infinite.
vector<int> maxi(k, numeric_limits<int>::min());
// Initialize cand array.
vector<int> cand(k);
// For each element of the prefix_sum array do:
// compute the cand array.
// take k maximum values from maxi and cand
// using maxmerge function.
// insert prefix_sum[i] to mini array if needed
// using insertMini function.
for (int i = 0; i < n; i++) {
for (int j = 0; j < k; j++) {
if(pre_sum[i] < 0 && mini[j]==numeric_limits<int>::max())
cand[j]=(-pre_sum[i])-mini[j];
else cand[j] = pre_sum[i] - mini[j];
}
maxMerge(maxi, cand);
insertMini(mini, pre_sum[i]);
}
// Elements of maxi array is the k
// maximum overlapping sub-array sums.
// Print out the elements of maxi array.
for (int ele : maxi)
cout << ele << " ";
cout << endl;
}
// Driver Program
int main()
{
// Test case 1
vector<int> arr1 = { 4, -8, 9, -4, 1, -8, -1, 6 };
int k1 = 4;
kMaxOvSubArray(arr1, k1);
// Test case 2
vector<int> arr2 = { -2, -3, 4, -1, -2, 1, 5, -3 };
int k2 = 3;
kMaxOvSubArray(arr2, k2);
return 0;
}
Python3
# Python program to find out k maximum sum of
# overlapping sub-arrays
# Function to compute prefix-sum of the input array
def prefix_sum(arr, n):
pre_sum = list()
pre_sum.append(arr[0])
for i in range(1, n):
pre_sum.append(pre_sum[i-1] + arr[i])
return pre_sum
# Update maxi by k maximum values from maxi and cand
def maxMerge(maxi, cand):
# Here cand and maxi arrays are in non-increasing
# order beforehand. Now, j is the index of the
# next cand element and i is the index of next
# maxi element. Traverse through maxi array.
# If cand[j] > maxi[i] insert cand[j] at the ith
# position in the maxi array and remove the minimum
# element of the maxi array i.e. the last element
# and increase j by 1 i.e. take the next element
# from cand.
k = len(maxi)
j = 0
for i in range(k):
if (cand[j] > maxi[i]):
maxi.insert(i, cand[j])
del maxi[-1]
j += 1
# Insert prefix_sum[i] to mini array if needed
def insertMini(mini, pre_sum):
# Traverse the mini array from left to right.
# If prefix_sum[i] is less than any element
# then insert prefix_sum[i] at that position
# and delete maximum element of the mini array
# i.e. the rightmost element from the array.
k = len(mini)
for i in range(k):
if (pre_sum < mini[i]):
mini.insert(i, pre_sum)
del mini[-1]
break
# Function to compute k maximum overlapping sub-array sums
def kMaxOvSubArray(arr, k):
n = len(arr)
# Compute the prefix sum of the input array.
pre_sum = prefix_sum(arr, n)
# Set all the elements of mini as + infinite
# except 0th. Set the 0th element as 0\.
mini = [float('inf') for i in range(k)]
mini[0] = 0
# Set all the elements of maxi as -infinite.
maxi = [-float('inf') for i in range(k)]
# Initialize cand array.
cand = [0 for i in range(k)]
# For each element of the prefix_sum array do:
# compute the cand array.
# take k maximum values from maxi and cand
# using maxmerge function.
# insert prefix_sum[i] to mini array if needed
# using insertMini function.
for i in range(n):
for j in range(k):
cand[j] = pre_sum[i] - mini[j]
maxMerge(maxi, cand)
insertMini(mini, pre_sum[i])
# Elements of maxi array is the k
# maximum overlapping sub-array sums.
# Print out the elements of maxi array.
for ele in maxi:
print(ele, end = ' ')
print('')
# Driver Program
# Test case 1
arr1 = [4, -8, 9, -4, 1, -8, -1, 6]
k1 = 4
kMaxOvSubArray(arr1, k1)
# Test case 2
arr2 = [-2, -3, 4, -1, -2, 1, 5, -3]
k2 = 3
kMaxOvSubArray(arr2, k2)
输出：
9 6 6 5
7 6 5
时间复杂度：insertMini和maxMerge函数以O(k)时间运行，并且需要O(k)时间来更新cand数组。 我们执行此过程n次。 因此，总时间复杂度为O(k * n)。
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