Dada una array arr[] que consta de N enteros positivos y un entero K , la tarea es igualar la suma de todos los subarreglos de longitud K reemplazando el número mínimo de elementos de la array con cualquier entero.
Ejemplos:
Entrada: arr[] = {3, 4, 3, 5, 6}, K = 2
Salida: 2
Explicación:
Operación 1: Reemplazar arr[3] por 4 modifica arr[] a {3, 4, 3, 4, 6}.
Operación 2: Reemplazar arr[4] por 3 modifica arr[] a {3, 4, 3, 4, 3}.
Todos los subarreglos de longitud 2 son {{3, 4}, {4, 3}, {3, 4}, {4, 3}}. La suma de todos estos subarreglos es 7. Por lo tanto, el número mínimo de operaciones requeridas es 2.Entrada: arr[] = {1, 2, 3, 1, 2}, K = 3
Salida: 0
Explicación: Todos los subarreglos de longitud 3 son {{1, 2, 3}, {2, 3, 1}, { 3, 1, 2}}. Dado que todos estos subarreglos tienen una suma de 6, el número de operaciones requeridas es 0.
Enfoque: La idea se basa en la observación de que todos los subarreglos tendrán la misma suma, cuando todos los elementos separados por la distancia K sean iguales.
Por lo tanto, el problema se puede resolver contando la frecuencia de elementos separados por una distancia K y encontrar el número que aparece el máximo de veces. Siga los pasos a continuación para resolver el problema:
- Inicialice una variable y almacene el resultado requerido.
- Iterar en el rango [0, K-1] usando la variable i
- Cree un mapa , freq para almacenar la frecuencia de los elementos separados por una distancia K a partir de i .
- Recorre el mapa y encuentra el elemento que aparece el máximo número de veces.
- Nuevamente, recorra el mapa y si el elemento no es igual al elemento máximo que ocurre, agregue su frecuencia a ans .
- Imprime el valor de ans como resultado.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find minimum number of
// operations required to make sum of
// all subarrays of size K equal
void findMinOperations(int arr[],
int N, int K)
{
// Stores number of operations
int operations = 0;
// Iterate in the range [0, K-1]
for (int i = 0; i < K; i++) {
// Stores frequency of elements
// separated by distance K
unordered_map<int, int> freq;
for (int j = i; j < N; j += K)
freq[arr[j]]++;
// Stores maximum frequency
// and corresponding element
int max1 = 0, num;
// Find max frequency element
// and its frequency
for (auto x : freq) {
if (x.second > max1) {
max1 = x.second;
num = x.first;
}
}
// Update the number of operations
for (auto x : freq) {
if (x.first != num)
operations += x.second;
}
}
// Print the result
cout << operations;
}
// Driver Code
int main()
{
// Given Input
int arr[] = { 3, 4, 3, 5, 6 };
int K = 2;
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
findMinOperations(arr, N, K);
return 0;
}
Java
// Java program for the above approach
import java.lang.*;
import java.util.*;
class GFG
{
// Function to find minimum number of
// operations required to make sum of
// all subarrays of size K equal
static void findMinOperations(int arr[],
int N, int K)
{
// Stores number of operations
int operations = 0;
// Iterate in the range [0, K-1]
for (int i = 0; i < K; i++) {
// Stores frequency of elements
// separated by distance K
Map<Integer, Integer> freq=new HashMap<>();
for (int j = i; j < N; j += K)
freq.put(arr[j], freq.getOrDefault(arr[j],0)+1);
// Stores maximum frequency
// and corresponding element
int max1 = 0, num=-1;
// Find max frequency element
// and its frequency
for (Map.Entry<Integer,Integer> x : freq.entrySet()) {
if (x.getValue() > max1) {
max1 = x.getValue();
num = x.getKey();
}
}
// Update the number of operations
for ( Map.Entry<Integer,Integer> x : freq.entrySet()) {
if (x.getKey() != num)
operations += x.getValue();
}
}
// Print the result
System.out.print(operations);
}
// Driver code
public static void main(String[] args)
{
// Given Input
int arr[] = { 3, 4, 3, 5, 6 };
int K = 2;
int N = arr.length;
// Function Call
findMinOperations(arr, N, K);
}
}
// This code is contributed by offbeat
Python3
# python 3 program for the above approach
# Function to find minimum number of
# operations required to make sum of
# all subarrays of size K equal
def findMinOperations(arr, N, K):
# Stores number of operations
operations = 0
# Iterate in the range [0, K-1]
for i in range(K):
# Stores frequency of elements
# separated by distance K
freq = {}
for j in range(i,N,K):
if arr[j] in freq:
freq[arr[j]] += 1
else:
freq[arr[j]] = 1
# Stores maximum frequency
# and corresponding element
max1 = 0
num = 0
# Find max frequency element
# and its frequency
for key,value in freq.items():
if (value > max1):
max1 = value
num = key
# Update the number of operations
for key,value in freq.items():
if (key != num):
operations += value
# Print the result
print(operations)
# Driver Code
if __name__ == '__main__':
# Given Input
arr = [3, 4, 3, 5, 6]
K = 2
N = len(arr)
# Function Call
findMinOperations(arr, N, K)
# This code is contributed by ipg2016107.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find minimum number of
// operations required to make sum of
// all subarrays of size K equal
static void findMinOperations(int []arr,
int N, int K)
{
// Stores number of operations
int operations = -1;
// Iterate in the range [0, K-1]
for(int i = 0; i < K; i++)
{
// Stores frequency of elements
// separated by distance K
Dictionary<int,
int> freq = new Dictionary<int,
int>();
for(int j = i; j < N; j += K)
{
if (freq.ContainsKey(arr[j]))
freq[arr[j]]++;
else
freq.Add(arr[j], 1);
}
// Stores maximum frequency
// and corresponding element
int max1 = -1, num = 0;
// Find max frequency element
// and its frequency
foreach(KeyValuePair<int, int> entry in freq)
{
if (entry.Key > max1)
{
max1 = entry.Value;
num = entry.Key;
}
}
// Update the number of operations
foreach(KeyValuePair<int, int> entry in freq)
{
if (entry.Key != num)
operations += entry.Value;
}
}
// Print the result
Console.Write(operations);
}
// Driver Code
public static void Main()
{
// Given Input
int []arr = { 3, 4, 3, 5, 6 };
int K = 2;
int N = arr.Length;
// Function Call
findMinOperations(arr, N, K);
}
}
// This code is contributed by SURENDRA_GANGWAR
Javascript
<script>
// JavaScript program for the above approach
// Function to find minimum number of
// operations required to make sum of
// all subarrays of size K equal
function findMinOperations(arr, N, K)
{
// Stores number of operations
var operations = 0;
var i,j;
// Iterate in the range [0, K-1]
for (i = 0; i < K; i++) {
// Stores frequency of elements
// separated by distance K
var freq = new Map();
for(j = i; j < N; j += K){
if(freq.has(arr[j]))
freq.set(arr[j], freq.get(arr[j])+1);
else
freq.set(arr[j],1);
}
// Stores maximum frequency
// and corresponding element
var max1 = 0, num;
// Find max frequency element
// and its frequency
for (const [key, value] of freq.entries()) {
if (value > max1) {
max1 = value;
num = key;
}
}
// Update the number of operations
for (const [key, value] of freq.entries()) {
if (key != num)
operations += value;
}
}
// Print the result
document.write(operations);
}
// Driver Code
// Given Input
var arr = [3, 4, 3, 5, 6];
var K = 2;
var N = arr.length;
// Function Call
findMinOperations(arr, N, K);
</script>
Producción:
2
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por pawanharwani11 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA