给定一个由N 个整数和一个整数D组成的数组arr[] ，任务是找到最小整数T ，使得整个数组可以从给定数组中划分为最多D个子数组，总和最大为T 。
例子：
Input: D = 5, arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 15
Explanation:
If T = 15, then 5 subarrays {{1, 2, 3, 4, 5}, {6, 7}, {8}, {9}, {10}}
Input: D = 2, arr[] = {1, 1, 1, 1, 1}
Output: 3
Explanation:
If T = 3, then the 2 partitions are {{1, 1, 1}, {1, 1}}
朴素的方法：这个想法是检查[max(element), sum(element)]范围内T 的所有可能值是否最多可能有 D 个分区。
时间复杂度： O(N*R)
辅助空间： O(1)
高效的方法：想法是使用二分搜索来优化上述方法。请按照以下步骤解决问题：
- 考虑R = [ max(element), sum(element) ]范围内的T 。
- 如果中位数T最多可以生成D 个分区，则检查小于T的中位数。
- 否则，检查是否存在大于当前中值T 的中值。
- 最后返回T的可能值。
下面是上述方法的实现：
C++
// C++ Program for the above approach
#include
using namespace std;
// Funtion to check if the array
// can be partitioned into atmost d
// subarray with sum atmost T
bool possible(int T, int arr[], int n, int d)
{
// Initial partition
int partition = 1;
// Current sum
int total = 0;
for (int i = 0; i < n; i++) {
total = total + arr[i];
// If current sum exceeds T
if (total > T) {
// Create a new partition
partition = partition + 1;
total = arr[i];
// If count of partitions
// exceed d
if (partition > d) {
return false;
}
}
}
return true;
}
// Function to find the minimum
// possible value of T
void calcT(int n, int d, int arr[])
{
// Stores the maximum and
// total sum of elements
int mx = -1, sum = 0;
for (int i = 0; i < n; i++) {
// Maximum element
mx = max(mx, arr[i]);
// Sum of all elements
sum = sum + arr[i];
}
int lb = mx;
int ub = sum;
while (lb < ub) {
// Calculate median T
int T_mid = lb + (ub - lb) / 2;
// If atmost D partitions possible
if (possible(T_mid, arr, n, d) == true) {
// Check for smaller T
ub = T_mid;
}
// Otherwise
else {
// Check for larger T
lb = T_mid + 1;
}
}
// Print the minimum T required
cout << lb << endl;
}
// Driver Code
int main()
{
int d = 2;
int arr[] = { 1, 1, 1, 1, 1 };
int n = sizeof arr / sizeof arr[0];
// Function call
calcT(n, d, arr);
return 0;
} Java
// Java program for the above approach
import java.util.*;
import java.io.*;
class GFG{
// Function to check if the array
// can be partitioned into atmost d
// subarray with sum atmost T
public static boolean possible(int T, int arr[],
int n, int d)
{
// Initial partition
int partition = 1;
// Current sum
int total = 0;
for(int i = 0; i < n; i++)
{
total = total + arr[i];
// If current sum exceeds T
if (total > T)
{
// Create a new partition
partition = partition + 1;
total = arr[i];
// If count of partitions
// exceed d
if (partition > d)
{
return false;
}
}
}
return true;
}
// Function to find the minimum
// possible value of T
public static void calcT(int n, int d,
int arr[])
{
// Stores the maximum and
// total sum of elements
int mx = -1, sum = 0;
for(int i = 0; i < n; i++)
{
// Maximum element
mx = Math.max(mx, arr[i]);
// Sum of all elements
sum = sum + arr[i];
}
int lb = mx;
int ub = sum;
while (lb < ub)
{
// Calculate median T
int T_mid = lb + (ub - lb) / 2;
// If atmost D partitions possible
if (possible(T_mid, arr, n, d) == true)
{
// Check for smaller T
ub = T_mid;
}
// Otherwise
else
{
// Check for larger T
lb = T_mid + 1;
}
}
// Print the minimum T required
System.out.println(lb);
}
// Driver code
public static void main(String args[])
{
int d = 2;
int arr[] = { 1, 1, 1, 1, 1 };
int n = arr.length;
// Function call
calcT(n, d, arr);
}
}
// This code is contributed by decodingPython3
# Python3 program for the above approach
# Function to check if the array
# can be partitioned into atmost d
# subarray with sum atmost T
def possible(T, arr, n, d):
# Initial partition
partition = 1;
# Current sum
total = 0;
for i in range(n):
total = total + arr[i];
# If current sum exceeds T
if (total > T):
# Create a new partition
partition = partition + 1;
total = arr[i];
# If count of partitions
# exceed d
if (partition > d):
return False;
return True;
# Function to find the minimum
# possible value of T
def calcT(n, d, arr):
# Stores the maximum and
# total sum of elements
mx = -1; sum = 0;
for i in range(n):
# Maximum element
mx = max(mx, arr[i]);
# Sum of all elements
sum = sum + arr[i];
lb = mx;
ub = sum;
while (lb < ub):
# Calculate median T
T_mid = lb + (ub - lb) // 2;
# If atmost D partitions possible
if (possible(T_mid, arr, n, d) == True):
# Check for smaller T
ub = T_mid;
# Otherwise
else:
# Check for larger T
lb = T_mid + 1;
# Print the minimum T required
print(lb);
# Driver code
if __name__ == '__main__':
d = 2;
arr = [ 1, 1, 1, 1, 1 ];
n = len(arr);
# Function call
calcT(n, d, arr);
# This code is contributed by Rajput-JiC#
// C# program for the above approach
using System;
class GFG{
// Function to check if the array
// can be partitioned into atmost d
// subarray with sum atmost T
public static bool possible(int T, int []arr,
int n, int d)
{
// Initial partition
int partition = 1;
// Current sum
int total = 0;
for(int i = 0; i < n; i++)
{
total = total + arr[i];
// If current sum exceeds T
if (total > T)
{
// Create a new partition
partition = partition + 1;
total = arr[i];
// If count of partitions
// exceed d
if (partition > d)
{
return false;
}
}
}
return true;
}
// Function to find the minimum
// possible value of T
public static void calcT(int n, int d,
int []arr)
{
// Stores the maximum and
// total sum of elements
int mx = -1, sum = 0;
for(int i = 0; i < n; i++)
{
// Maximum element
mx = Math.Max(mx, arr[i]);
// Sum of all elements
sum = sum + arr[i];
}
int lb = mx;
int ub = sum;
while (lb < ub)
{
// Calculate median T
int T_mid = lb + (ub - lb) / 2;
// If atmost D partitions possible
if (possible(T_mid, arr, n, d) == true)
{
// Check for smaller T
ub = T_mid;
}
// Otherwise
else
{
// Check for larger T
lb = T_mid + 1;
}
}
// Print the minimum T required
Console.WriteLine(lb);
}
// Driver code
public static void Main(String []args)
{
int d = 2;
int []arr = { 1, 1, 1, 1, 1 };
int n = arr.Length;
// Function call
calcT(n, d, arr);
}
}
// This code is contributed by 29AjayKumarJavascript
输出
3时间复杂度： O( N*log(sum) )
辅助空间： O(1)
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