给定一个大小为N的数组arr[] ，任务是找到数组索引的计数，以便从这些索引中删除一个元素使偶数索引和奇数索引数组元素的总和相等。
例子：
Input: arr[] = { 2, 1, 6, 4 }
Output: 1
Explanation:
Removing arr[1] from the array modifies arr[] to { 2, 6, 4 } such that, arr[0] + arr[2] = arr[1].
Therefore, the required output is 1.
Input: arr[] = { 1, 1, 1 }
Output: 3
Explanation:
Removing arr[0] from the given array modifies arr[] to { 1, 1 } such that arr[0] = arr[1]
Removing arr[1] from the given array modifies arr[] to { 1, 1 } such that arr[0] = arr[1]
Removing arr[2] from the given array modifies arr[] to { 1, 1 } such that arr[0] = arr[1]
Therefore, the required output is 3.
朴素的方法：解决这个问题的最简单的方法是遍历数组，对于每个数组元素，检查从数组中删除元素是否使偶数索引和奇数索引数组元素的总和相等。如果发现为真，则增加计数。最后，打印计数。
时间复杂度： O(N 2 )
辅助空间： O(1)
有效方法：上述方法可以基于以下观察进行优化：从给定数组中删除任何元素会使后续元素的偶数索引为奇数，而后继元素的奇数索引为偶数。请按照以下步骤解决问题：
- 初始化两个变量，比如evenSum和oddSum ，分别存储给定数组的奇数索引和偶数索引元素的总和。
- 使用变量i遍历数组。
- 删除数组的每个第i个元素，并根据上述观察更新剩余的偶数索引元素和奇数索引元素的总和。检查总和是否相等。如果发现为真，则增加计数。
- 最后，打印获得的计数。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to count array indices
// whose removal makes sum of odd and
// even indexed elements equal
int cntIndexesToMakeBalance(int arr[], int n)
{
// If size of the array is 1
if (n == 1) {
return 1;
}
// If size of the array is 2
if (n == 2)
return 0;
// Stores sum of even-indexed
// elements of the given array
int sumEven = 0;
// Stores sum of odd-indexed
// elements of the given array
int sumOdd = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// If i is an even number
if (i % 2 == 0) {
// Update sumEven
sumEven += arr[i];
}
// If i is an odd number
else {
// Update sumOdd
sumOdd += arr[i];
}
}
// Stores sum of even-indexed
// array elements till i-th index
int currOdd = 0;
// Stores sum of odd-indexed
// array elements till i-th index
int currEven = arr[0];
// Stores count of indices whose
// removal makes sum of odd and
// even indexed elements equal
int res = 0;
// Stores sum of even-indexed elements
// after removing the i-th element
int newEvenSum = 0;
// Stores sum of odd-indexed elements
// after removing the i-th element
int newOddSum = 0;
// Traverse the array
for (int i = 1; i < n - 1; i++) {
// If i is an odd number
if (i % 2) {
// Update currOdd
currOdd += arr[i];
// Update newEvenSum
newEvenSum = currEven + sumOdd
- currOdd;
// Update newOddSum
newOddSum = currOdd + sumEven
- currEven - arr[i];
}
// If i is an even number
else {
// Update currEven
currEven += arr[i];
// Update newOddSum
newOddSum = currOdd + sumEven
- currEven;
// Update newEvenSum
newEvenSum = currEven + sumOdd
- currOdd - arr[i];
}
// If newEvenSum is equal to newOddSum
if (newEvenSum == newOddSum) {
// Increase the count
res++;
}
}
// If sum of even-indexed and odd-indexed
// elements is equal by removing the first element
if (sumOdd == sumEven - arr[0]) {
// Increase the count
res++;
}
// If length of the array
// is an odd number
if (n % 2 == 1) {
// If sum of even-indexed and odd-indexed
// elements is equal by removing the last element
if (sumOdd == sumEven - arr[n - 1]) {
// Increase the count
res++;
}
}
// If length of the array
// is an even number
else {
// If sum of even-indexed and odd-indexed
// elements is equal by removing the last element
if (sumEven == sumOdd - arr[n - 1]) {
// Increase the count
res++;
}
}
return res;
}
// Driver Code
int main()
{
int arr[] = { 1, 1, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << cntIndexesToMakeBalance(arr, n);
return 0;
} Java
// Java program to implement
// the above approach
class GFG {
// Function to count array indices
// whose removal makes sum of odd and
// even indexed elements equal
static int cntIndexesToMakeBalance(int arr[], int n)
{
// If size of the array is 1
if (n == 1)
{
return 1;
}
// If size of the array is 2
if (n == 2)
return 0;
// Stores sum of even-indexed
// elements of the given array
int sumEven = 0;
// Stores sum of odd-indexed
// elements of the given array
int sumOdd = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// If i is an even number
if (i % 2 == 0)
{
// Update sumEven
sumEven += arr[i];
}
// If i is an odd number
else
{
// Update sumOdd
sumOdd += arr[i];
}
}
// Stores sum of even-indexed
// array elements till i-th index
int currOdd = 0;
// Stores sum of odd-indexed
// array elements till i-th index
int currEven = arr[0];
// Stores count of indices whose
// removal makes sum of odd and
// even indexed elements equal
int res = 0;
// Stores sum of even-indexed elements
// after removing the i-th element
int newEvenSum = 0;
// Stores sum of odd-indexed elements
// after removing the i-th element
int newOddSum = 0;
// Traverse the array
for (int i = 1; i < n - 1; i++)
{
// If i is an odd number
if (i % 2 != 0)
{
// Update currOdd
currOdd += arr[i];
// Update newEvenSum
newEvenSum = currEven + sumOdd
- currOdd;
// Update newOddSum
newOddSum = currOdd + sumEven
- currEven - arr[i];
}
// If i is an even number
else
{
// Update currEven
currEven += arr[i];
// Update newOddSum
newOddSum = currOdd + sumEven
- currEven;
// Update newEvenSum
newEvenSum = currEven + sumOdd
- currOdd - arr[i];
}
// If newEvenSum is equal to newOddSum
if (newEvenSum == newOddSum)
{
// Increase the count
res++;
}
}
// If sum of even-indexed and odd-indexed
// elements is equal by removing the first element
if (sumOdd == sumEven - arr[0])
{
// Increase the count
res++;
}
// If length of the array
// is an odd number
if (n % 2 == 1)
{
// If sum of even-indexed and odd-indexed
// elements is equal by removing the last element
if (sumOdd == sumEven - arr[n - 1])
{
// Increase the count
res++;
}
}
// If length of the array
// is an even number
else
{
// If sum of even-indexed and odd-indexed
// elements is equal by removing the last element
if (sumEven == sumOdd - arr[n - 1])
{
// Increase the count
res++;
}
}
return res;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 1, 1, 1 };
int n = arr.length;
System.out.println(cntIndexesToMakeBalance(arr, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program to implement
# the above approach
# Function to count array indices
# whose removal makes sum of odd and
# even indexed elements equal
def cntIndexesToMakeBalance(arr, n):
# If size of the array is 1
if (n == 1):
return 1
# If size of the array is 2
if (n == 2):
return 0
# Stores sum of even-indexed
# elements of the given array
sumEven = 0
# Stores sum of odd-indexed
# elements of the given array
sumOdd = 0
# Traverse the array
for i in range(n):
# If i is an even number
if (i % 2 == 0):
# Update sumEven
sumEven += arr[i]
# If i is an odd number
else:
# Update sumOdd
sumOdd += arr[i]
# Stores sum of even-indexed
# array elements till i-th index
currOdd = 0
# Stores sum of odd-indexed
# array elements till i-th index
currEven = arr[0]
# Stores count of indices whose
# removal makes sum of odd and
# even indexed elements equal
res = 0
# Stores sum of even-indexed elements
# after removing the i-th element
newEvenSum = 0
# Stores sum of odd-indexed elements
# after removing the i-th element
newOddSum = 0
# Traverse the array
for i in range(1, n - 1):
# If i is an odd number
if (i % 2):
# Update currOdd
currOdd += arr[i]
# Update newEvenSum
newEvenSum = (currEven + sumOdd -
currOdd)
# Update newOddSum
newOddSum = (currOdd + sumEven -
currEven - arr[i])
# If i is an even number
else:
# Update currEven
currEven += arr[i]
# Update newOddSum
newOddSum = (currOdd + sumEven -
currEven)
# Update newEvenSum
newEvenSum = (currEven + sumOdd -
currOdd - arr[i])
# If newEvenSum is equal to newOddSum
if (newEvenSum == newOddSum):
# Increase the count
res += 1
# If sum of even-indexed and odd-indexed
# elements is equal by removing the first
# element
if (sumOdd == sumEven - arr[0]):
# Increase the count
res += 1
# If length of the array
# is an odd number
if (n % 2 == 1):
# If sum of even-indexed and odd-indexed
# elements is equal by removing the last
# element
if (sumOdd == sumEven - arr[n - 1]):
# Increase the count
res += 1
# If length of the array
# is an even number
else:
# If sum of even-indexed and odd-indexed
# elements is equal by removing the last
# element
if (sumEven == sumOdd - arr[n - 1]):
# Increase the count
res += 1
return res
# Driver Code
if __name__ == "__main__" :
arr = [ 1, 1, 1 ]
n = len(arr)
print(cntIndexesToMakeBalance(arr, n))
# This code is contributed by AnkitRai01C#
// C# program to implement
// the above approach
using System;
class GFG {
// Function to count array indices
// whose removal makes sum of odd and
// even indexed elements equal
static int cntIndexesToMakeBalance(int[] arr, int n)
{
// If size of the array is 1
if (n == 1)
{
return 1;
}
// If size of the array is 2
if (n == 2)
return 0;
// Stores sum of even-indexed
// elements of the given array
int sumEven = 0;
// Stores sum of odd-indexed
// elements of the given array
int sumOdd = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// If i is an even number
if (i % 2 == 0)
{
// Update sumEven
sumEven += arr[i];
}
// If i is an odd number
else
{
// Update sumOdd
sumOdd += arr[i];
}
}
// Stores sum of even-indexed
// array elements till i-th index
int currOdd = 0;
// Stores sum of odd-indexed
// array elements till i-th index
int currEven = arr[0];
// Stores count of indices whose
// removal makes sum of odd and
// even indexed elements equal
int res = 0;
// Stores sum of even-indexed elements
// after removing the i-th element
int newEvenSum = 0;
// Stores sum of odd-indexed elements
// after removing the i-th element
int newOddSum = 0;
// Traverse the array
for (int i = 1; i < n - 1; i++)
{
// If i is an odd number
if (i % 2 != 0)
{
// Update currOdd
currOdd += arr[i];
// Update newEvenSum
newEvenSum = currEven + sumOdd
- currOdd;
// Update newOddSum
newOddSum = currOdd + sumEven
- currEven - arr[i];
}
// If i is an even number
else
{
// Update currEven
currEven += arr[i];
// Update newOddSum
newOddSum = currOdd + sumEven
- currEven;
// Update newEvenSum
newEvenSum = currEven + sumOdd
- currOdd - arr[i];
}
// If newEvenSum is equal to newOddSum
if (newEvenSum == newOddSum)
{
// Increase the count
res++;
}
}
// If sum of even-indexed and odd-indexed
// elements is equal by removing the first element
if (sumOdd == sumEven - arr[0])
{
// Increase the count
res++;
}
// If length of the array
// is an odd number
if (n % 2 == 1)
{
// If sum of even-indexed and odd-indexed
// elements is equal by removing the last element
if (sumOdd == sumEven - arr[n - 1])
{
// Increase the count
res++;
}
}
// If length of the array
// is an even number
else
{
// If sum of even-indexed and odd-indexed
// elements is equal by removing the last element
if (sumEven == sumOdd - arr[n - 1])
{
// Increase the count
res++;
}
}
return res;
}
// Drivers Code
public static void Main ()
{
int[] arr = { 1, 1, 1 };
int n = arr.Length;
Console.WriteLine(cntIndexesToMakeBalance(arr, n));
}
}
// This code is contributed by susmitakundugoaldangaJavascript
输出：
3时间复杂度： O(N)
辅助空间： O(1)
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