修改给定数组以使奇数和偶数索引元素的总和相同 _奇数和偶数索引元素

本文概述

C ++
Java
Python3
C#

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，使奇数和偶数下标的元素之和相等。任务是打印修改后的数组。
注意:N总是偶数。可能有多个结果，打印任何一个。
例子：

【修改给定数组以使奇数和偶数索引元素的总和相同】输入：arr [] = {1, 1, 1, 0}
输出：1 1
说明：对于给定的数组arr [] = {1, 1, 1, 0}在奇数位置的元素之和, Odd_Sum = arr [ 1] + arr [3] = 1 +1 =2。偶数位置的元素之和, Even_Sum = arr [2] + arr [4] = 1 + 0 =1。由于Odd_Sum和Even_Sum不相等, 因此删除一些使它们相等的元素。
删除arr [3]和arr [4]后, 数组变为arr [] = {1, 1}, 以使奇数索引和偶数索引处的元素之和相等。
输入：arr [] = {1, 0}
输出：0
说明：对于给定的数组arr [] = {1, 0}在奇数位置的元素之和, Odd_Sum = arr [1] = 0 = 0。偶数位置上的元素的平均, Even_Sum = 1 =1。由于Odd_Sum和Even_Sum不相等, 请删除一些元素以使其相等。删除arr [0]后, 数组变为arr [] = {0}, 以使奇数索引和偶数索引处的元素之和相等。

方法：这个想法是为了数数1s和0s在给定数组中, 然后通过以下步骤使结果总和相等：

数数0s和1s在给定的数组中arr []并将它们存储在变量中count_0和count_1分别。
如果在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。
If计数_0≥N / 2, 然后删除所有1s并打印所有零count_0次数。
否则, 如果计数_0 < N / 2, 通过删除所有0s, 可以使偶数和奇数索引的总和与以下条件相同：
- Ifcount_1是奇数, 然后打印1作为新数组的元素(计数_1 – 1)次数。
- 否则, 打印1作为新数组的元素count_1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 < bits/stdc++.h> 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 = 0for 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# 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

输出如下：