给定N个整数和两个整数S和M的数组arr [] ，任务是通过将大小为S的任何子数组增加1到M次来最大化最小数组元素。
例子：
Input: arr[] = {1, 2, 3, 4, 5, 6}, S = 2, M = 3
Output: 3
Explanation:
Below are the operations performed:
Operation 1: Select subarray {1, 2} and after increment, array arr[] becomes = {2, 3, 3, 4, 5, 6}.
Operation 2: Select subarray {2, 3} and after increment, array arr[] becomes = {3, 4, 3, 4, 5, 6}.
Operation 3: Select subarray {3, 4} and after increment, array arr[] becomes = {4, 5, 3, 4, 5, 6}.
After the above operations, the minimum element of the array is 3.
Input: arr[] = {3, 5, 2, 7, 3}, S = 3, M = 3
Output: 4
Explanation:
Below are the operations performed:
Operation 1: Select subarray {3, 5, 2} and after increment, array arr[] becomes = {4, 6, 3, 7, 3}.
Operation 2: Select subarray {4, 6, 3} and after increment, array arr[] becomes = {5, 7, 4, 7, 3}.
Operation 3: Select subarray {4, 7, 3} and after increment, array arr[] becomes = {5, 7, 5, 8, 4}.
After the above operations, the minimum element of the array is 4.
方法：这个想法是找到数组M的最小元素次，然后从该最小元素开始将大小为S的子数组增加1 。请按照以下步骤解决问题：
- 遍历数组M次，对于每次迭代，请执行以下操作：
- 找到数组arr []的最小元素。令最小元素的第一个索引为idx 。
- 将当前的最小元素增加1 。
- 现在将两个指针leftIdx作为idx – 1和rightIdx作为idx + 1 。
- 如果leftIdx处的元素小于rightIdx处的元素，则将A [leftIndex]递增1，并将leftIndex递减1 。否则，将A [rightIndex]增大1并将rightIndex增大1 。继续此步骤，直到处理(S – 1)个元素。
- 在上述迭代之后，打印更新后的数组的最小元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to return index of minimum
// element in the array
int min(int a[], int n)
{
// Initialize a[0] as minValue
int minIndex = 0, minValue = a[0], i;
// Traverse the array
for (i = 1; i < n; i++) {
// If a[i] < existing minValue
if (a[i] < minValue) {
minValue = a[i];
minIndex = i;
}
}
// Return the minimum index
return minIndex;
}
// Function that maximize the minimum
// element of array after incrementing
// subarray of size S by 1, M times
int maximizeMin(int A[], int N,
int S, int M)
{
int minIndex, left, right, i, j;
// Iterating through the array
// for M times
for (i = 0; i < M; i++) {
// Find minimum element index
minIndex = min(A, N);
// Increment the minimum value
A[minIndex]++;
// Storing the left index
// and right index
left = minIndex - 1;
right = minIndex + 1;
// Incrementing S - 1 minumum
// elements to the left and
// right of minValue
for (j = 0; j < S - 1; j++) {
// Reached extreme left
if (left == -1)
A[right++]++;
// Reached extreme right
else if (right == N)
A[left--]++;
else {
// Left value is minimum
if (A[left] < A[right])
A[left--]++;
// Right value is minimum
else
A[right++]++;
}
}
}
// Find the minValue in A[] after
// M operations
minIndex = min(A, N);
// Return the minimum value
return A[minIndex];
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int N = sizeof(arr) / sizeof(arr[0]);
int S = 2, M = 3;
// Function Call
cout << maximizeMin(arr, N, S, M);
return 0;
} Java
// Java program for the
// above approach
import java.util.*;
class solution{
// Function to return index
// of minimum element in the
// array
static int min1(int a[], int n)
{
// Initialize a[0] as
// minValue
int minIndex = 0,
minValue = a[0], i;
// Traverse the array
for (i = 1; i < n; i++)
{
// If a[i] < existing
// minValue
if (a[i] < minValue)
{
minValue = a[i];
minIndex = i;
}
}
// Return the minimum index
return minIndex;
}
// Function that maximize the minimum
// element of array after incrementing
// subarray of size S by 1, M times
static int maximizeMin(int A[], int N,
int S, int M)
{
int minIndex, left, right, i, j;
// Iterating through the
// array or M times
for (i = 0; i < M; i++)
{
// Find minimum element
// index
minIndex = min1(A, N);
// Increment the minimum
// value
A[minIndex]++;
// Storing the left index
// and right index
left = minIndex - 1;
right = minIndex + 1;
// Incrementing S - 1 minumum
// elements to the left and
// right of minValue
for (j = 0; j < S - 1; j++)
{
// Reached extreme left
if (left == -1)
A[right++]++;
// Reached extreme right
else if (right == N)
A[left--]++;
else
{
// Left value is minimum
if (A[left] < A[right])
A[left--]++;
// Right value is minimum
else
A[right++]++;
}
}
}
// Find the minValue in A[] after
// M operations
minIndex = min1(A, N);
// Return the minimum value
return A[minIndex];
}
// Driver Code
public static void main(String args[])
{
int []arr = {1, 2, 3,
4, 5, 6};
int N = arr.length;
int S = 2, M = 3;
// Function Call
System.out.print(maximizeMin(arr, N, S, M));
}
}
// This code is contributed by SURENDRA_GANGWARPython3
# Python3 program for the above approach
# Function to return index of minimum
# element in the array
def min(a, n):
# Initialize a[0] as minValue
minIndex = 0
minValue = a[0]
# Traverse the array
for i in range(1, n):
# If a[i] < existing minValue
if (a[i] < minValue):
minValue = a[i]
minIndex = i
# Return the minimum index
return minIndex
# Function that maximize the minimum
# element of array after incrementing
# subarray of size S by 1, M times
def maximizeMin(A, N, S, M):
minIndex, left, right = 0, 0, 0
# Iterating through the array
# for M times
for i in range(M):
# Find minimum element index
minIndex = min(A, N)
# Increment the minimum value
A[minIndex] += 1
# Storing the left index
# and right index
left = minIndex - 1
right = minIndex + 1
# Incrementing S - 1 minumum
# elements to the left and
# right of minValue
for j in range(S - 1):
# Reached extreme left
if (left == -1):
A[right] += 1
right += 1
# Reached extreme right
elif (right == N):
A[left] += 1
left -= 1
else:
# Left value is minimum
if (A[left] < A[right]):
A[left] += 1
left -= 1
# Right value is minimum
else:
A[right] += 1
right += 1
# Find the minValue in A[] after
# M operations
minIndex = min(A, N)
# Return the minimum value
return A[minIndex]
# Driver Code
if __name__ == '__main__':
arr = [ 1, 2, 3, 4, 5, 6 ]
N = len(arr)
S = 2
M = 3
#Function Call
print(maximizeMin(arr, N, S, M))
# This code is contributed by mohit kumar 29C#
// C# program for the
// above approach
using System;
class GFG{
// Function to return index
// of minimum element in the
// array
static int min1(int[] a,
int n)
{
// Initialize a[0] as
// minValue
int minIndex = 0,
minValue = a[0], i;
// Traverse the array
for (i = 1; i < n; i++)
{
// If a[i] < existing
// minValue
if (a[i] < minValue)
{
minValue = a[i];
minIndex = i;
}
}
// Return the minimum
// index
return minIndex;
}
// Function that maximize the
// minimum element of array
// after incrementing subarray
// of size S by 1, M times
static int maximizeMin(int[] A, int N,
int S, int M)
{
int minIndex, left, right, i, j;
// Iterating through the
// array or M times
for (i = 0; i < M; i++)
{
// Find minimum element
// index
minIndex = min1(A, N);
// Increment the minimum
// value
A[minIndex]++;
// Storing the left index
// and right index
left = minIndex - 1;
right = minIndex + 1;
// Incrementing S - 1 minumum
// elements to the left and
// right of minValue
for (j = 0; j < S - 1; j++)
{
// Reached extreme left
if (left == -1)
A[right++]++;
// Reached extreme right
else if (right == N)
A[left--]++;
else
{
// Left value is minimum
if (A[left] < A[right])
A[left--]++;
// Right value is minimum
else
A[right++]++;
}
}
}
// Find the minValue in A[] after
// M operations
minIndex = min1(A, N);
// Return the minimum value
return A[minIndex];
}
// Driver code
static void Main()
{
int[] arr = {1, 2, 3,
4, 5, 6};
int N = arr.Length;
int S = 2, M = 3;
// Function Call
Console.Write(maximizeMin(arr, N,
S, M));
}
}
// This code is contributed by divyeshrabadiya07输出：
3时间复杂度： O(M * N)
辅助空间： O(1)