给定一个大小为N的二进制数组arr[] ,从数组中最多删除N/2 个元素,使得奇数和偶数索引处的元素总和变得相等。任务是打印修改后的数组。
注意: N 总是偶数。可能有多个可能的结果,打印其中任何一个。
例子:
Input: arr[] = {1, 1, 1, 0}
Output: 1 1
Explanation:
For the given array arr[] = {1, 1, 1, 0}
The sum of elements at odd position, Odd_Sum = arr[1] + arr[3] = 1 + 1 = 2.
The sum of elements at even position, Even_Sum = arr[2] + arr[4] = 1 + 0 = 1.
Since Odd_Sum & Even_Sum are not equal, remove some elements to make them equal.
After removing arr[3] and arr[4] the array become arr[] = {1, 1} such that sum of elements at odd indices and even indices are equal.
Input: arr[] = {1, 0}
Output: 0
Explanation:
For the given array arr[] = {1, 0}
The sum of elements at odd position, Odd_Sum = arr[1] = 0 = 0.
The sum of elements at even position, Even_Sum = 1 = 1.
Since Odd_Sum & Even_Sum are not equal, remove some elements to make them equal.
After removing arr[0] the array become arr[] = {0} such that sum of elements at odd indices and even indices are equal.
方法:想法是计算给定数组中1和0的数量,然后通过以下步骤使结果和相等:
- 计算给定数组arr[]中0和1的数量,并将它们分别存储在变量中,例如count_0和count_1 。
- 如果奇数和偶数索引处的元素之和相等,则无需删除任何数组元素。打印原始数组作为答案。
- 如果count_0 ≥ N/2 ,则删除所有1并打印所有零count_0次。
- 否则,如果count_0 < N/2 ,通过删除所有0 s,可以根据以下条件使偶数和奇数索引的总和相同:
- 如果count_1 为奇数,则将1作为新数组的元素打印(count_1 – 1)次。
- 否则,打印1作为新数组count_1的元素次数。
- 根据上述条件打印结果数组。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to modify array to make
// sum of odd and even indexed
// elements equal
void makeArraySumEqual(int a[], int N)
{
// Stores the count of 0s, 1s
int count_0 = 0, count_1 = 0;
// Stores sum of odd and even
// indexed elements respectively
int odd_sum = 0, even_sum = 0;
for (int i = 0; i < N; i++) {
// Count 0s
if (a[i] == 0)
count_0++;
// Count 1s
else
count_1++;
// Calculate odd_sum and even_sum
if ((i + 1) % 2 == 0)
even_sum += a[i];
else if ((i + 1) % 2 > 0)
odd_sum += a[i];
}
// If both are equal
if (odd_sum == even_sum) {
// Print the original array
for (int i = 0; i < N; i++)
printf("%d ", a[i]);
}
// Otherwise
else {
if (count_0 >= N / 2) {
// Print all the 0s
for (int i = 0; i < count_0; i++)
printf("0 ");
}
else {
// For checking even or odd
int is_Odd = count_1 % 2;
// Update total count of 1s
count_1 -= is_Odd;
// Print all 1s
for (int i = 0; i < count_1; i++)
printf("1 ");
}
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 1, 1, 0 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
makeArraySumEqual(arr, N);
return 0;
}
Java
// Java program for the above approach
class GFG {
// Function to modify array to make
// sum of odd and even indexed
// elements equal
static void makeArraySumEqual(int a[], int N)
{
// Stores the count of 0s, 1s
int count_0 = 0, count_1 = 0;
// Stores sum of odd and even
// indexed elements respectively
int odd_sum = 0, even_sum = 0;
for (int i = 0; i < N; i++) {
// Count 0s
if (a[i] == 0)
count_0++;
// Count 1s
else
count_1++;
// Calculate odd_sum and even_sum
if ((i + 1) % 2 == 0)
even_sum += a[i];
else if ((i + 1) % 2 > 0)
odd_sum += a[i];
}
// If both are equal
if (odd_sum == even_sum) {
// Print the original array
for (int i = 0; i < N; i++)
System.out.print(a[i] + " ");
}
else
{
if (count_0 >= N / 2) {
// Print all the 0s
for (int i = 0; i < count_0; i++)
System.out.print("0 ");
}
else {
// For checking even or odd
int is_Odd = count_1 % 2;
// Update total count of 1s
count_1 -= is_Odd;
// Print all 1s
for (int i = 0; i < count_1; i++)
System.out.print("1 ");
}
}
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1, 1, 1, 0 };
int N = arr.length;
// Function Call
makeArraySumEqual(arr, N);
}
}
Python3
# Python3 program for the above approach
# Function to modify array to make
# sum of odd and even indexed
# elements equal
def makeArraySumEqual(a, N):
# Stores the count of 0s, 1s
count_0 = 0
count_1 = 0
# Stores sum of odd and even
# indexed elements respectively
odd_sum = 0
even_sum = 0
for i in range(N):
# Count 0s
if (a[i] == 0):
count_0 += 1
# Count 1s
else:
count_1 += 1
# Calculate odd_sum and even_sum
if ((i + 1) % 2 == 0):
even_sum += a[i]
elif ((i + 1) % 2 > 0):
odd_sum += a[i]
# If both are equal
if (odd_sum == even_sum):
# Print the original array
for i in range(N):
print(a[i], end = " ")
# Otherwise
else:
if (count_0 >= N / 2):
# Print all the 0s
for i in range(count_0):
print("0", end = " ")
else:
# For checking even or odd
is_Odd = count_1 % 2
# Update total count of 1s
count_1 -= is_Odd
# Print all 1s
for i in range(count_1):
print("1", end = " ")
# Driver Code
# Given array arr[]
arr = [ 1, 1, 1, 0 ]
N = len(arr)
# Function call
makeArraySumEqual(arr, N)
# This code is contributed by code_hunt
C#
// C# program for
// the above approach
using System;
class GFG{
// Function to modify array to make
// sum of odd and even indexed
// elements equal
static void makeArraySumEqual(int []a,
int N)
{
// Stores the count of 0s, 1s
int count_0 = 0, count_1 = 0;
// Stores sum of odd and even
// indexed elements respectively
int odd_sum = 0, even_sum = 0;
for (int i = 0; i < N; i++)
{
// Count 0s
if (a[i] == 0)
count_0++;
// Count 1s
else
count_1++;
// Calculate odd_sum and even_sum
if ((i + 1) % 2 == 0)
even_sum += a[i];
else if ((i + 1) % 2 > 0)
odd_sum += a[i];
}
// If both are equal
if (odd_sum == even_sum)
{
// Print the original array
for (int i = 0; i < N; i++)
Console.Write(a[i] + " ");
}
else
{
if (count_0 >= N / 2)
{
// Print all the 0s
for (int i = 0; i < count_0; i++)
Console.Write("0 ");
}
else
{
// For checking even or odd
int is_Odd = count_1 % 2;
// Update total count of 1s
count_1 -= is_Odd;
// Print all 1s
for (int i = 0; i < count_1; i++)
Console.Write("1 ");
}
}
}
// Driver Code
public static void Main(String[] args)
{
// Given array []arr
int []arr = {1, 1, 1, 0};
int N = arr.Length;
// Function Call
makeArraySumEqual(arr, N);
}
}
// This code is contributed by 29AjayKumar
Javascript
1 1
时间复杂度: O(N)
辅助空间: O(1)
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