给定正整数数组，请替换数组中的每个元素，以使数组中相邻元素之间的差小于或等于给定目标。我们需要最小化调整成本，即新值和旧值之间的差之和。我们基本上需要最小化＆Sum; | A [i] –新的[i] |其中0≤i≤n-1，n是A []的大小，而new []是相邻差小于或等于目标的数组。
假设数组的所有元素都小于常数M = 100。
例子：
Input: arr = [1, 3, 0, 3], target = 1
Output: Minimum adjustment cost is 3
Explanation: One of the possible solutions
is [2, 3, 2, 3]
Input: arr = [2, 3, 2, 3], target = 1
Output: Minimum adjustment cost is 0
Explanation: All adjacent elements in the input
array are already less than equal to given target
Input: arr = [55, 77, 52, 61, 39, 6,
25, 60, 49, 47], target = 10
Output: Minimum adjustment cost is 75
Explanation: One of the possible solutions is
[55, 62, 52, 49, 39, 29, 30, 40, 49, 47]
为了使调整成本最小化|| A [i] –新的[i] |对于数组中的所有索引i，| A [i] –新的[i] |应该尽可能接近零。此外，| A [i] –新的[i + 1]] | ≤目标。
这个问题可以通过动态编程解决。
令dp [i] [j]定义将A [i]更改为j时的最小调整成本，则DP关系定义为–
dp[i][j] = min{dp[i - 1][k]} + |j - A[i]|
for all k's such that |k - j| ≤ target
在这里，0≤i≤n和0≤j≤M，其中n是数组中元素的数量，M =100。我们必须考虑所有k，使得max(j – target，0)≤k≤min(M， j +目标)
最后，对于所有0≤j≤M，阵列的最小调整成本为min {dp [n – 1] [j]}。
以下是上述想法的实现–
C++
// C++ program to find minimum adjustment cost of an array
#include
using namespace std;
#define M 100
// Function to find minimum adjustment cost of an array
int minAdjustmentCost(int A[], int n, int target)
{
// dp[i][j] stores minimal adjustment cost on changing
// A[i] to j
int dp[n][M + 1];
// handle first element of array separately
for (int j = 0; j <= M; j++)
dp[0][j] = abs(j - A[0]);
// do for rest elements of the array
for (int i = 1; i < n; i++)
{
// replace A[i] to j and calculate minimal adjustment
// cost dp[i][j]
for (int j = 0; j <= M; j++)
{
// initialize minimal adjustment cost to INT_MAX
dp[i][j] = INT_MAX;
// consider all k such that k >= max(j - target, 0) and
// k <= min(M, j + target) and take minimum
for (int k = max(j-target,0); k <= min(M,j+target); k++)
dp[i][j] = min(dp[i][j], dp[i - 1][k] + abs(A[i] - j));
}
}
// return minimum value from last row of dp table
int res = INT_MAX;
for (int j = 0; j <= M; j++)
res = min(res, dp[n - 1][j]);
return res;
}
// Driver Program to test above functions
int main()
{
int arr[] = {55, 77, 52, 61, 39, 6, 25, 60, 49, 47};
int n = sizeof(arr) / sizeof(arr[0]);
int target = 10;
cout << "Minimum adjustment cost is "
<< minAdjustmentCost(arr, n, target) << endl;
return 0;
} Java
// Java program to find minimum adjustment cost of an array
import java.io.*;
import java.util.*;
class GFG
{
public static int M = 100;
// Function to find minimum adjustment cost of an array
static int minAdjustmentCost(int A[], int n, int target)
{
// dp[i][j] stores minimal adjustment cost on changing
// A[i] to j
int[][] dp = new int[n][M + 1];
// handle first element of array separately
for (int j = 0; j <= M; j++)
dp[0][j] = Math.abs(j - A[0]);
// do for rest elements of the array
for (int i = 1; i < n; i++)
{
// replace A[i] to j and calculate minimal adjustment
// cost dp[i][j]
for (int j = 0; j <= M; j++)
{
// initialize minimal adjustment cost to INT_MAX
dp[i][j] = Integer.MAX_VALUE;
// consider all k such that k >= max(j - target, 0) and
// k <= min(M, j + target) and take minimum
int k = Math.max(j-target,0);
for ( ; k <= Math.min(M,j+target); k++)
dp[i][j] = Math.min(dp[i][j], dp[i - 1][k] +
Math.abs(A[i] - j));
}
}
// return minimum value from last row of dp table
int res = Integer.MAX_VALUE;
for (int j = 0; j <= M; j++)
res = Math.min(res, dp[n - 1][j]);
return res;
}
// Driver program
public static void main (String[] args)
{
int arr[] = {55, 77, 52, 61, 39, 6, 25, 60, 49, 47};
int n = arr.length;
int target = 10;
System.out.println("Minimum adjustment cost is "
+minAdjustmentCost(arr, n, target));
}
}
// This code is contributed by Pramod KumarPython3
# Python3 program to find minimum
# adjustment cost of an array
M = 100
# Function to find minimum
# adjustment cost of an array
def minAdjustmentCost(A, n, target):
# dp[i][j] stores minimal adjustment
# cost on changing A[i] to j
dp = [[0 for i in range(M + 1)]
for i in range(n)]
# handle first element
# of array separately
for j in range(M + 1):
dp[0][j] = abs(j - A[0])
# do for rest elements
# of the array
for i in range(1, n):
# replace A[i] to j and
# calculate minimal adjustment
# cost dp[i][j]
for j in range(M + 1):
# initialize minimal adjustment
# cost to INT_MAX
dp[i][j] = 100000000
# consider all k such that
# k >= max(j - target, 0) and
# k <= min(M, j + target) and
# take minimum
for k in range(max(j - target, 0),
min(M, j + target) + 1):
dp[i][j] = min(dp[i][j], dp[i - 1][k] +
abs(A[i] - j))
# return minimum value from
# last row of dp table
res = 10000000
for j in range(M + 1):
res = min(res, dp[n - 1][j])
return res
# Driver Code
arr= [55, 77, 52, 61, 39,
6, 25, 60, 49, 47]
n = len(arr)
target = 10
print("Minimum adjustment cost is",
minAdjustmentCost(arr, n, target),
sep = ' ')
# This code is contributed
# by sahilshelangiaC#
// C# program to find minimum adjustment
// cost of an array
using System;
class GFG {
public static int M = 100;
// Function to find minimum adjustment
// cost of an array
static int minAdjustmentCost(int []A, int n,
int target)
{
// dp[i][j] stores minimal adjustment
// cost on changing A[i] to j
int[,] dp = new int[n,M + 1];
// handle first element of array
// separately
for (int j = 0; j <= M; j++)
dp[0,j] = Math.Abs(j - A[0]);
// do for rest elements of the array
for (int i = 1; i < n; i++)
{
// replace A[i] to j and calculate
// minimal adjustment cost dp[i][j]
for (int j = 0; j <= M; j++)
{
// initialize minimal adjustment
// cost to INT_MAX
dp[i,j] = int.MaxValue;
// consider all k such that
// k >= max(j - target, 0) and
// k <= min(M, j + target) and
// take minimum
int k = Math.Max(j - target, 0);
for ( ; k <= Math.Min(M, j +
target); k++)
dp[i,j] = Math.Min(dp[i,j],
dp[i - 1,k]
+ Math.Abs(A[i] - j));
}
}
// return minimum value from last
// row of dp table
int res = int.MaxValue;
for (int j = 0; j <= M; j++)
res = Math.Min(res, dp[n - 1,j]);
return res;
}
// Driver program
public static void Main ()
{
int []arr = {55, 77, 52, 61, 39,
6, 25, 60, 49, 47};
int n = arr.Length;
int target = 10;
Console.WriteLine("Minimum adjustment"
+ " cost is "
+ minAdjustmentCost(arr, n, target));
}
}
// This code is contributed by Sam007.PHP
= max(j - target, 0) and
// k <= min(M, j + target) and
// take minimum
for($k = max($j - $target, 0);
$k <= min($M, $j + $target);
$k++)
$dp[$i][$j] = min($dp[$i][$j],
$dp[$i - 1][$k] +
abs($A[$i] - $j));
}
}
// return minimum value
// from last row of dp table
$res = PHP_INT_MAX;
for($j = 0; $j <= $M; $j++)
$res = min($res, $dp[$n - 1][$j]);
return $res;
}
// Driver Code
$arr = array(55, 77, 52, 61, 39,
6, 25, 60, 49, 47);
$n = count($arr);
$target = 10;
echo "Minimum adjustment cost is "
, minAdjustmentCost($arr, $n, $target);
// This code is contributed by anuj_67.
?>输出：
Minimum adjustment cost is 75