给定一个长度为 ‘N’ 的排序数组和一个整数 ‘K'(K
Input : arr[] = {1, 2, 3, 4}, k = 1
Output : 2
Let's consider all possible cases.
a) Remove 0th index: arr[] = {2, 3, 4}, ans = 2
b) Remove 1th index: arr[] = {1, 3, 4}, ans = 3
c) Remove 2th index: arr[] = {1, 2, 4}, ans = 3
d) Remove 3th index: arr[] = {1, 2, 3}, ans = 2
Minimum of them all is 2, thus answer = 2
Input : arr[] = {1, 2, 10}, k = 1
Output : 1方法 ：
从末端移除元素是减少总和值的唯一可能方法。
例如，让数组 = {1, 2, 3, 4}。如果删除了数组中的第二个元素，则总和保持与前一个相同，即等于 3。但是，如果删除第一个或最后一个元素，则总和减少到 2。
贪婪方法：在每一步中，删除使总和减少更多的元素。例如，让数组 = {1, 3, 9, 33}，33 被删除，因为它将总和从 32 减少到 8。
但是，这种贪婪的方法不适用于某些测试用例。例子。 arr[] = {1, 2, 100, 120, 140} 和 k = 2。这里，贪心方法的最终数组是 {1, 2, 100} 而最优数组是 {100, 120, 140}。
动态规划： DP的状态如下：
DP[l][r] 表示通过删除子数组 arr[l 到 r] 中所需数量的元素可以获得的最小总和。
因此，递推关系将是
DP[l][r] = min(DP[l][r-1], DP[l+1][r])下面是上述想法的 C++ 实现。
C++
// C++ implementation of the above approach.
#include
using namespace std;
#define N 100
#define INF 1000000
// states of DP
int dp[N][N];
bool vis[N][N];
// function to find minimum sum
int findSum(int* arr, int n, int k, int l, int r)
{
// base-case
if ((l) + (n - 1 - r) == k)
return arr[r] - arr[l];
// if state is solved before, return
if (vis[l][r])
return dp[l][r];
// marking the state as solved
vis[l][r] = 1;
// recurrence relation
return dp[l][r] = min(findSum(arr, n, k, l, r - 1),
findSum(arr, n, k, l + 1, r));
}
// driver function
int32_t main()
{
// input values
int arr[] = { 1, 2, 100, 120, 140 };
int k = 2;
int n = sizeof(arr) / sizeof(int);
// callin the required function;
cout << findSum(arr, n, k, 0, n - 1);
} Java
// Java implementation of the above approach.
class GFG
{
final static int N = 100 ;
final static int INF = 1000000 ;
// states of DP
static int dp[][] = new int[N][N];
static int vis[][] = new int[N][N];
// function to find minimum sum
static int findSum(int []arr, int n,
int k, int l, int r)
{
// base-case
if ((l) + (n - 1 - r) == k)
return arr[r] - arr[l];
// if state is solved before, return
if (vis[l][r] == 1)
return dp[l][r];
// marking the state as solved
vis[l][r] = 1;
// recurrence relation
dp[l][r] = Math.min(findSum(arr, n, k, l, r - 1),
findSum(arr, n, k, l + 1, r));
return dp[l][r] ;
}
// Driver function
public static void main (String[] args)
{
// input values
int arr[] = { 1, 2, 100, 120, 140 };
int k = 2;
int n = arr.length;
// calling the required function;
System.out.println(findSum(arr, n, k, 0, n - 1));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the above approach.
import numpy as np
N = 100
INF = 1000000
# states of DP
dp = np.zeros((N, N));
vis = np.zeros((N, N));
# function to find minimum sum
def findSum(arr, n, k, l, r) :
# base-case
if ((l) + (n - 1 - r) == k) :
return arr[r] - arr[l];
# if state is solved before, return
if (vis[l][r]) :
return dp[l][r];
# marking the state as solved
vis[l][r] = 1;
# recurrence relation
dp[l][r] = min(findSum(arr, n, k, l, r - 1),
findSum(arr, n, k, l + 1, r));
return dp[l][r]
# driver function
if __name__ == "__main__" :
# input values
arr = [ 1, 2, 100, 120, 140 ];
k = 2;
n = len(arr);
# calling the required function;
print(findSum(arr, n, k, 0, n - 1));
# This code is contributed by AnkitRai01C#
// C# implementation of the above approach.
using System;
class GFG
{
static int N = 100 ;
// states of DP
static int [,]dp = new int[N, N];
static int [,]vis = new int[N, N];
// function to find minimum sum
static int findSum(int []arr, int n,
int k, int l, int r)
{
// base-case
if ((l) + (n - 1 - r) == k)
return arr[r] - arr[l];
// if state is solved before, return
if (vis[l, r] == 1)
return dp[l, r];
// marking the state as solved
vis[l, r] = 1;
// recurrence relation
dp[l, r] = Math.Min(findSum(arr, n, k, l, r - 1),
findSum(arr, n, k, l + 1, r));
return dp[l, r] ;
}
// Driver function
public static void Main ()
{
// input values
int []arr = { 1, 2, 100, 120, 140 };
int k = 2;
int n = arr.Length;
// calling the required function;
Console.WriteLine(findSum(arr, n, k, 0, n - 1));
}
}
// This code is contributed by AnkitRai01Javascript
C++
//C++ implementation of the above approach.
#include
using namespace std;
// function to find minimum sum
int findSum(int* arr, int n, int k)
{
// variable to store final answer
// and initialising it with the values
// when 0 elements is removed from the left and
// K from the right.
int ans = arr[n - k - 1] - arr[0];
// loop to simulate removal of elements
for (int i = 1; i <= k; i++) {
//removing i elements from the left and and K-i elements
//from the right and updating the answer correspondingly
ans = min(arr[n - 1 - (k - i)] - arr[i], ans);
}
// returning final answer
return ans;
}
// driver function
int32_t main()
{
// input values
int arr[] = { 1, 2, 100, 120, 140 };
int k = 2;
int n = sizeof(arr) / sizeof(int);
// callin the required function;
cout << findSum(arr, n, k);
} Java
// Java implementation of the above approach.
class GFG
{
// function to find minimum sum
static int findSum(int []arr, int n, int k)
{
// variable to store final answer
// and initialising it with the values
// when 0 elements is removed from the left and
// K from the right.
int ans = arr[n - k - 1] - arr[0];
// loop to simulate removal of elements
for (int i = 1; i <= k; i++)
{
// removing i elements from the left and and K-i elements
// from the right and updating the answer correspondingly
ans = Math.min(arr[n - 1 - (k - i)] - arr[i], ans);
}
// returning final answer
return ans;
}
// Driver function
public static void main (String[] args)
{
// input values
int arr[] = { 1, 2, 100, 120, 140 };
int k = 2;
int n = arr.length;
// callin the required function;
System.out.println(findSum(arr, n, k));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the above approach.
# function to find minimum sum
def findSum(arr, n, k) :
# variable to store final answer
# and initialising it with the values
# when 0 elements is removed from the left and
# K from the right.
ans = arr[n - k - 1] - arr[0];
# loop to simulate removal of elements
for i in range(1, k + 1) :
# removing i elements from the left and and K-i elements
# from the right and updating the answer correspondingly
ans = min(arr[n - 1 - (k - i)] - arr[i], ans);
# returning final answer
return ans;
# Driver code
if __name__ == "__main__" :
# input values
arr = [ 1, 2, 100, 120, 140 ];
k = 2;
n = len(arr);
# calling the required function;
print(findSum(arr, n, k));
# This code is contributed by AnkitRai01C#
// C# implementation of the above approach.
using System;
class GFG
{
// function to find minimum sum
static int findSum(int []arr, int n, int k)
{
// variable to store final answer
// and initialising it with the values
// when 0 elements is removed from the left and
// K from the right.
int ans = arr[n - k - 1] - arr[0];
// loop to simulate removal of elements
for (int i = 1; i <= k; i++)
{
// removing i elements from the left and and K-i elements
// from the right and updating the answer correspondingly
ans = Math.Min(arr[n - 1 - (k - i)] - arr[i], ans);
}
// returning final answer
return ans;
}
// Driver function
public static void Main ()
{
// input values
int []arr = { 1, 2, 100, 120, 140 };
int k = 2;
int n = arr.Length;
// calling the required function;
Console.WriteLine(findSum(arr, n, k));
}
}
// This code is contributed by AnkitRai01Javascript
输出：
40时间复杂度： O(n^2)
注意：此问题也存在 O(N) 方法。但是上述方法也可以通过稍加修改来解决未排序数组的问题。
替代方法：
仅从左角和右角移除元素。因此，如果从左侧移除x 个元素，那么对于(0,K) 中的每个 x，都会从右侧移除Kx 个元素。
如果执行上述操作，差异之和将等于arr[N-(KX)-1] – arr[X] 。
在从 (0, K) 迭代 x 时，从获得的值中挑选最小值。
例子：
Input: arr[] = {1, 3, 7, 8, 13} ; k = 3
Output: 1
Explanation:
Looping from X = 0 to X = K;
1) X = 0 and K-X = 3
So 0 elements removed from left and 3 from right.
array will be {1, 3} and answer will be 3 - 1 = 2.
min = 2
2) X = 1 and K-X = 2
So 1 elements removed from left and 2 from right.
array will be {3, 7} and answer will be 7 - 3 = 4.
min = 2
3) X = 2 and K-X = 1
So 2 elements removed from left and 1 from right.
array will be {7, 8} and answer will be 8 - 7 = 1.
min = 1
4) X = 3 and K-X = 0
So 3 elements removed from left and 0 from right.
array will be {8, 13} and answer will be 13 - 8 = 5.
min = 1下面是上述方法的实现。
C++
//C++ implementation of the above approach.
#include
using namespace std;
// function to find minimum sum
int findSum(int* arr, int n, int k)
{
// variable to store final answer
// and initialising it with the values
// when 0 elements is removed from the left and
// K from the right.
int ans = arr[n - k - 1] - arr[0];
// loop to simulate removal of elements
for (int i = 1; i <= k; i++) {
//removing i elements from the left and and K-i elements
//from the right and updating the answer correspondingly
ans = min(arr[n - 1 - (k - i)] - arr[i], ans);
}
// returning final answer
return ans;
}
// driver function
int32_t main()
{
// input values
int arr[] = { 1, 2, 100, 120, 140 };
int k = 2;
int n = sizeof(arr) / sizeof(int);
// callin the required function;
cout << findSum(arr, n, k);
}
Java
// Java implementation of the above approach.
class GFG
{
// function to find minimum sum
static int findSum(int []arr, int n, int k)
{
// variable to store final answer
// and initialising it with the values
// when 0 elements is removed from the left and
// K from the right.
int ans = arr[n - k - 1] - arr[0];
// loop to simulate removal of elements
for (int i = 1; i <= k; i++)
{
// removing i elements from the left and and K-i elements
// from the right and updating the answer correspondingly
ans = Math.min(arr[n - 1 - (k - i)] - arr[i], ans);
}
// returning final answer
return ans;
}
// Driver function
public static void main (String[] args)
{
// input values
int arr[] = { 1, 2, 100, 120, 140 };
int k = 2;
int n = arr.length;
// callin the required function;
System.out.println(findSum(arr, n, k));
}
}
// This code is contributed by AnkitRai01
蟒蛇3
# Python3 implementation of the above approach.
# function to find minimum sum
def findSum(arr, n, k) :
# variable to store final answer
# and initialising it with the values
# when 0 elements is removed from the left and
# K from the right.
ans = arr[n - k - 1] - arr[0];
# loop to simulate removal of elements
for i in range(1, k + 1) :
# removing i elements from the left and and K-i elements
# from the right and updating the answer correspondingly
ans = min(arr[n - 1 - (k - i)] - arr[i], ans);
# returning final answer
return ans;
# Driver code
if __name__ == "__main__" :
# input values
arr = [ 1, 2, 100, 120, 140 ];
k = 2;
n = len(arr);
# calling the required function;
print(findSum(arr, n, k));
# This code is contributed by AnkitRai01
C＃
// C# implementation of the above approach.
using System;
class GFG
{
// function to find minimum sum
static int findSum(int []arr, int n, int k)
{
// variable to store final answer
// and initialising it with the values
// when 0 elements is removed from the left and
// K from the right.
int ans = arr[n - k - 1] - arr[0];
// loop to simulate removal of elements
for (int i = 1; i <= k; i++)
{
// removing i elements from the left and and K-i elements
// from the right and updating the answer correspondingly
ans = Math.Min(arr[n - 1 - (k - i)] - arr[i], ans);
}
// returning final answer
return ans;
}
// Driver function
public static void Main ()
{
// input values
int []arr = { 1, 2, 100, 120, 140 };
int k = 2;
int n = arr.Length;
// calling the required function;
Console.WriteLine(findSum(arr, n, k));
}
}
// This code is contributed by AnkitRai01
Javascript
输出：
40如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。