给定一个长度为N的数组arr[]和一个整数K 。在每个操作中，可以从数组中选择任何元素（比如arr[i] ），并且可以将其更改为arr[i] + 1或arr[i] – 1 。任务是找到对数组执行所需的最少操作次数，以使数组模 K 的每个值保持不变。
例子：
Input: arr[] = {4, 5, 8, 3, 12}, k =5
Output: 4
Explanation:
Operation 1: { 3, 5, 8, 3, 12 }, decrease 4 at index 0 by 1.
Operation 2: { 3, 4, 8, 3, 12 }, decrease 5 at index 1 by 1.
Operation 3: { 3, 3, 8, 3, 12 }, decrease 4 at index 1 by 1.
Operation 4: { 3, 3, 8, 3, 13 }, increase 12 at index 4 by 1.
The modulo of each number is equal to 3 and minimum steps required were 4.
Input: arr[] = {2, 35, 48, 23, 52}, k =3
Output: 2
Explanation:
Minimum number of steps required to make modulo of each number equal is 2.
方法：这个想法是使用散列来保留已获得的每个模的计数。
- 现在迭代 i 的每个值，范围为 0 <= i < k，并找到使所有数字的模相等所需的运算次数。
- 如果它小于获得的值比当前存储的值，则更新它。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum operations
// required to make the modulo of each
// element of the array equal to each other
int Find_min(set& diff_mod,
map count_mod, int k)
{
// Variable to store minimum
// operation required
int min_oprn = INT_MAX;
// To store operation required to
// make all modulo equal
int oprn = 0;
// Iterating through all
// possible modulo value
for (int x = 0; x < k; x++) {
oprn = 0;
// Iterating through all different
// modulo obtained so far
for (auto w : diff_mod) {
// Calculating oprn required
// to make all modulos equal
// to x
if (w != x) {
if (w == 0) {
// Checking the operations
// that will cost less
oprn += min(x, k - x)
* count_mod[w];
}
else {
// Check operation that
// will cost less
oprn += min(
abs(x - w),
k + x - w)
* count_mod[w];
}
}
}
// Update the minimum
// number of operations
if (oprn < min_oprn)
min_oprn = oprn;
}
// Returing the answer
return min_oprn;
}
// Function to store different modulos
int Cal_min(int arr[], int n, int k)
{
// Set to store all
// different modulo
set diff_mod;
// Map to store count
// of all different modulo
// obtained
map count_mod;
// Storing all different
// modulo count
for (int i = 0; i < n; i++) {
// Insert into the set
diff_mod.insert(arr[i] % k);
// Increment count
count_mod[arr[i] % k]++;
}
// Function call to return value of
// min oprn required
return Find_min(diff_mod, count_mod, k);
}
// Driver Code
int main()
{
int arr[] = { 2, 35, 48, 23, 52 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 3;
cout << Cal_min(arr, n, k);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the minimum operations
// required to make the modulo of each
// element of the array equal to each other
static int Find_min(HashSet diff_mod,
HashMap count_mod,
int k)
{
// Variable to store minimum
// operation required
int min_oprn = Integer.MAX_VALUE;
// To store operation required to
// make all modulo equal
int oprn = 0;
// Iterating through all
// possible modulo value
for(int x = 0; x < k; x++)
{
oprn = 0;
// Iterating through all different
// modulo obtained so far
for(int w : diff_mod)
{
// Calculating oprn required
// to make all modulos equal
// to x
if (w != x)
{
if (w == 0)
{
// Checking the operations
// that will cost less
oprn += Math.min(x, k - x) *
count_mod.get(w);
}
else
{
// Check operation that
// will cost less
oprn += Math.min(Math.abs(x - w),
k + x - w) *
count_mod.get(w);
}
}
}
// Update the minimum
// number of operations
if (oprn < min_oprn)
min_oprn = oprn;
}
// Returing the answer
return min_oprn;
}
// Function to store different modulos
static int Cal_min(int arr[], int n, int k)
{
// Set to store all
// different modulo
HashSet diff_mod = new HashSet<>();
// Map to store count
// of all different modulo
// obtained
HashMap count_mod = new HashMap<>();
// Storing all different
// modulo count
for(int i = 0; i < n; i++)
{
// Insert into the set
diff_mod.add(arr[i] % k);
// Increment count
count_mod.put(arr[i] % k,
count_mod.getOrDefault(arr[i] % k, 0) + 1);
}
// Function call to return value of
// min oprn required
return Find_min(diff_mod, count_mod, k);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 2, 35, 48, 23, 52 };
int n = arr.length;
int k = 3;
System.out.print(Cal_min(arr, n, k));
}
}
// This code is contributed by jrishabh99 Python3
# Python3 program for
# the above approach
import sys
from collections import defaultdict
# Function to find the minimum operations
# required to make the modulo of each
# element of the array equal to each other
def Find_min(diff_mod,
count_mod, k):
# Variable to store minimum
# operation required
min_oprn = sys.maxsize
# To store operation required to
# make all modulo equal
oprn = 0
# Iterating through all
# possible modulo value
for x in range (k):
oprn = 0
# Iterating through all different
# modulo obtained so far
for w in diff_mod:
# Calculating oprn required
# to make all modulos equal
# to x
if (w != x):
if (w == 0):
# Checking the operations
# that will cost less
oprn += (min(x, k - x) *
count_mod[w])
else:
# Check operation that
# will cost less
oprn += (min(abs(x - w),
k + x - w) *
count_mod[w])
# Update the minimum
# number of operations
if (oprn < min_oprn):
min_oprn = oprn
# Returing the answer
return min_oprn
# Function to store different modulos
def Cal_min(arr, n, k):
# Set to store all
# different modulo
diff_mod = set([])
# Map to store count
# of all different modulo
# obtained
count_mod = defaultdict (int)
# Storing all different
# modulo count
for i in range (n):
# Insert into the set
diff_mod.add(arr[i] % k)
# Increment count
count_mod[arr[i] % k] += 1
# Function call to return value of
# min oprn required
return Find_min(diff_mod, count_mod, k)
# Driver Code
if __name__ == "__main__":
arr = [2, 35, 48, 23, 52]
n = len(arr)
k = 3
print( Cal_min(arr, n, k))
# This code is contributed by ChitranayalC#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the minimum operations
// required to make the modulo of each
// element of the array equal to each other
static int Find_min(HashSet diff_mod,
Dictionary count_mod,
int k)
{
// Variable to store minimum
// operation required
int min_oprn = int.MaxValue;
// To store operation required to
// make all modulo equal
int oprn = 0;
// Iterating through all
// possible modulo value
for(int x = 0; x < k; x++)
{
oprn = 0;
// Iterating through all different
// modulo obtained so far
foreach(int w in diff_mod)
{
// Calculating oprn required
// to make all modulos equal
// to x
if (w != x)
{
if (w == 0)
{
// Checking the operations
// that will cost less
oprn += Math.Min(x, k - x) *
count_mod[w];
}
else
{
// Check operation that
// will cost less
oprn += Math.Min(Math.Abs(x - w),
k + x - w) *
count_mod[w];
}
}
}
// Update the minimum
// number of operations
if (oprn < min_oprn)
min_oprn = oprn;
}
// Returing the answer
return min_oprn;
}
// Function to store different modulos
static int Cal_min(int []arr, int n, int k)
{
// Set to store all
// different modulo
HashSet diff_mod = new HashSet();
// Map to store count
// of all different modulo
// obtained
Dictionary count_mod = new Dictionary();
// Storing all different
// modulo count
for(int i = 0; i < n; i++)
{
// Insert into the set
diff_mod.Add(arr[i] % k);
// Increment count
if(count_mod.ContainsKey((arr[i] % k)))
count_mod[arr[i] % k] = count_mod[(arr[i] % k)]+1;
else
count_mod.Add(arr[i] % k, 1);
}
// Function call to return value of
// min oprn required
return Find_min(diff_mod, count_mod, k);
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 2, 35, 48, 23, 52 };
int n = arr.Length;
int k = 3;
Console.Write(Cal_min(arr, n, k));
}
}
// This code is contributed by Amit Katiyar Javascript
输出：
2时间复杂度： O(N*K)