给定一个由 n 个正整数组成的数组。找到要添加到数组中的索引之一的最小正元素，以便可以将其划分为两个等和的连续子数组。输出要添加的最小元素和要添加的位置。如果可能有多个位置，则返回至少一个。
例子：
Input : arr[] = { 10, 1, 2, 3, 4 }
Output : 0 0
Explanation: The array can already be partitioned into two contiguous subarrays i.e. {10} and {1, 2, 3, 4} with equal sums. So we need to add 0 at any position. (least position is 0)
Input : arr[] = { 5, 4 }
Output : 1 1
Explanation: We need to add 1 to 4 so that two sub arrays are {5},{5} hence output is 1 1 (minimum element and it’s position where it is to be added.
先决条件：前缀总和
方法一(简单)：一个天真的方法是计算从(arr[0]到arr[i])和(arr[i+1]到arr[n-1])的元素之和，并在每一步找到这两个总和之间的差异，最小差异将给出要添加的元素。
方法二(高效) ：仔细分析，我们可以使用前缀和和后缀和的概念。计算它们并找到 prefixSum[i](其中 prefixSum[i] 是数组元素到第 i 个位置的总和)和 suffixSum[i + 1](其中 suffixSum[i + 1] 是数组的总和)之间的绝对差从第 (i + 1) 个位置到最后一个位置的元素)。
例如
arr 10 1 2 3 4
prefixSum 10 11 13 16 20
suffixSum 20 10 9 7 4
Absolute difference between suffixSum[i + 1] and prefixSum[i]
10 - 10 = 0 // minimum
11 - 9 = 1
13 - 7 = 6
16 - 4 = 12
Thus, minimum element is 0, for position,
if prefixSum[i] is greater than suffixSum[i + 1], then position is "i + 1".
else it's is "i".下面是上述方法的实现。
C++
// CPP Program to find the minimum element to
// be added such that the array can be partitioned
// into two contiguous subarrays with equal sums
#include
using namespace std;
// Structure to store the minimum element
// and its position
struct data {
int element;
int position;
};
struct data findMinElement(int arr[], int n)
{
struct data result;
// initialize prefix and suffix sum arrays with 0
int prefixSum[n] = { 0 };
int suffixSum[n] = { 0 };
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++) {
// add current element to Sum
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
suffixSum[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
// add current element to Sum
suffixSum[i] = suffixSum[i + 1] + arr[i];
}
// initialize the minimum element to be a large value
int min = suffixSum[0];
int pos;
for (int i = 0; i < n - 1; i++) {
// check for the minimum absolute difference
// between current prefix sum and the next
// suffix sum element
if (abs(suffixSum[i + 1] - prefixSum[i]) < min) {
min = abs(suffixSum[i + 1] - prefixSum[i]);
// if prefixsum has a greater value then position
// is the next element, else it's the same element.
if (suffixSum[i + 1] < prefixSum[i]) pos = i + 1;
else pos = i;
}
}
// return the data in struct.
result.element = min;
result.position = pos;
return result;
}
// Driver Code
int main()
{
int arr[] = { 10, 1, 2, 3, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
struct data values;
values = findMinElement(arr, n);
cout << "Minimum element : " << values.element
<< endl << "Position : " << values.position;
return 0;
} Java
// Java Program to find the minimum element to
// be added such that the array can be partitioned
// into two contiguous subarrays with equal sums
import java.util.*;
class GFG
{
// Structure to store the minimum element
// and its position
static class data
{
int element;
int position;
};
static data findMinElement(int arr[], int n)
{
data result=new data();
// initialize prefix and suffix sum arrays with 0
int []prefixSum = new int[n];
int []suffixSum = new int[n];
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++)
{
// add current element to Sum
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
suffixSum[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--)
{
// add current element to Sum
suffixSum[i] = suffixSum[i + 1] + arr[i];
}
// initialize the minimum element to be a large value
int min = suffixSum[0];
int pos=0;
for (int i = 0; i < n - 1; i++)
{
// check for the minimum absolute difference
// between current prefix sum and the next
// suffix sum element
if (Math.abs(suffixSum[i + 1] - prefixSum[i]) < min)
{
min = Math.abs(suffixSum[i + 1] - prefixSum[i]);
// if prefixsum has a greater value then position
// is the next element, else it's the same element.
if (suffixSum[i + 1] < prefixSum[i]) pos = i + 1;
else pos = i;
}
}
// return the data in struct.
result.element = min;
result.position = pos;
return result;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 10, 1, 2, 3, 4 };
int n = arr.length;
data values;
values = findMinElement(arr, n);
System.out.println("Minimum element : " + values.element
+ "\nPosition : " + values.position);
}
}
// This code is contributed by Rajput-JiPython3
# Python Program to find the minimum element to
# be added such that the array can be partitioned
# into two contiguous subarrays with equal sums
# Class to store the minimum element
# and its position
class Data:
def __init__(self):
self.element = -1
self.position = -1
def findMinElement(arr, n):
result = Data()
# initialize prefix and suffix sum arrays with 0
prefixSum = [0]*n
suffixSum = [0]*n
prefixSum[0] = arr[0]
for i in range(1, n):
# add current element to Sum
prefixSum[i] = prefixSum[i-1] + arr[i]\
suffixSum[n-1] = arr[n-1]
for i in range(n-2, -1, -1):
# add current element to Sum
suffixSum[i] = suffixSum[i+1] + arr[i]
# initialize the minimum element to be a large value
mini = suffixSum[0]
pos = 0
for i in range(n-1):
# check for the minimum absolute difference
# between current prefix sum and the next
# suffix sum element
if abs(suffixSum[i+1]-prefixSum[i]) < mini:
mini = abs(suffixSum[i+1] - prefixSum[i])
# if prefixsum has a greater value then position
# is the next element, else it's the same element.
if suffixSum[i+1] < prefixSum[i]:
pos = i+1
else:
pos = i
# return the data in class.
result.element = mini
result.position = pos
return result
# Driver Code
if __name__ == "__main__":
arr = [10, 1, 2, 3, 4]
n = len(arr)
values = Data()
values = findMinElement(arr, n)
print("Minimum element :", values.element, "\nPosition :", values.position)
# This code is contributed by
# sanjeev2552C#
// C# Program to find the minimum element
// to be added such that the array can be
// partitioned into two contiguous subarrays
// with equal sums
using System;
class GFG
{
// Structure to store the minimum element
// and its position
public class data
{
public int element;
public int position;
};
static data findMinElement(int []arr, int n)
{
data result = new data();
// initialize prefix and suffix
// sum arrays with 0
int []prefixSum = new int[n];
int []suffixSum = new int[n];
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++)
{
// add current element to Sum
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
suffixSum[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--)
{
// add current element to Sum
suffixSum[i] = suffixSum[i + 1] + arr[i];
}
// initialize the minimum element
// to be a large value
int min = suffixSum[0];
int pos = 0;
for (int i = 0; i < n - 1; i++)
{
// check for the minimum absolute difference
// between current prefix sum and the next
// suffix sum element
if (Math.Abs(suffixSum[i + 1] -
prefixSum[i]) < min)
{
min = Math.Abs(suffixSum[i + 1] -
prefixSum[i]);
// if prefixsum has a greater value then position
// is the next element, else it's the same element.
if (suffixSum[i + 1] < prefixSum[i])
pos = i + 1;
else
pos = i;
}
}
// return the data in struct.
result.element = min;
result.position = pos;
return result;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 10, 1, 2, 3, 4 };
int n = arr.Length;
data values;
values = findMinElement(arr, n);
Console.WriteLine("Minimum element : " +
values.element +
"\nPosition : " +
values.position);
}
}
// This code is contributed by PrinciRaj1992Javascript
输出：
Minimum element : 0
Position : 0如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。