找到循环数组中前缀和总是非负的索引
给定一个由 N 个整数组成的圆形数组 arr[] ，任务是找到圆形数组的起始索引，使得来自该索引的前缀和总是非负的。如果没有这样的索引，则打印-1”。
示例:
输入: arr[] = {3，-6，7，-1，-4，5，-1} 输出: 2 解释: 考虑计算给定循环数组前缀和的索引 2，然后前缀和由{9，3，7，6，2，7，6}给出
输入: arr[] = {3，-5，-1} 输出: -1
方法:给定的问题可以基于以下观察来解决:
- 如果数组元素的和为负，则不存在从该点开始前缀和为非负的索引。
- 否则，该索引是在具有最小前缀和的索引之后的索引。
按照以下步骤解决问题:
- 初始化一个变量，比如说 sum 为 0 ，存储数组元素的 sum。
- 初始化一个变量，比如说中的为 0 ，存储循环遍历的起始索引。
- 初始化一个变量，说 min 为 INT_MAX ，存储数组的最小前缀和 arr[] 。
- 遍历给定数组并执行以下步骤:
- 将总和的值更新为总和和当前元素arr【I】的总和。
- 如果和的值小于分，则将分更新为和，中的为 (i + 1) 。
- 如果数组的和为负，则打印 -1 。否则，打印(单位为% N) 的值作为最终可能的索引。
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 find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
int startingPoint(int A[], int N)
{
// Stores the sum of the array
int sum = 0;
// Stores the starting index
int in = 0;
// Stores the minimum prefix
// sum of A[0..i]
int min = INT_MAX;
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
// Update the value of sum
sum += A[i];
// If sum is less than min
if (sum < min) {
// Update the min as the
// value of prefix sum
min = sum;
// Update in
in = i + 1;
}
}
// Otherwise, no such index is
// possible
if (sum < 0) {
return -1;
}
return in % N;
}
// Driver Code
int main()
{
int arr[] = { 3, -6, 7, -4, -4, 6, -1 };
int N = (sizeof(arr) / sizeof(arr[0]));
cout << startingPoint(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
static int startingPoint(int A[], int N)
{
// Stores the sum of the array
int sum = 0;
// Stores the starting index
int in = 0;
// Stores the minimum prefix
// sum of A[0..i]
int min = Integer.MAX_VALUE;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Update the value of sum
sum += A[i];
// If sum is less than min
if (sum < min)
{
// Update the min as the
// value of prefix sum
min = sum;
// Update in
in = i + 1;
}
}
// Otherwise, no such index is
// possible
if (sum < 0)
{
return -1;
}
return in % N;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 3, -6, 7, -4, -4, 6, -1 };
int N = arr.length;
System.out.print(startingPoint(arr, N));
}
}
// This code is contributed by Kingash
Python 3
# Python3 program for the above approach
# Function to find the starting index
# of the given circular array
# prefix sum array is non negative
import sys
def startingPoint(A, N):
# Stores the sum of the array
sum = 0
# Stores the starting index
startingindex = 0
# Stores the minimum prefix
# sum of A[0..i]
min = sys.maxsize
# Traverse the array
for i in range(0, N):
# Update the value of sum
sum += A[i]
# If sum is less than minimum
if (sum < min):
# Update the min as
# the value of prefix sum
min = sum
# Update starting index
startingindex = i + 1
# Otherwise no such index is possible
if (sum < 0):
return -1
return startingindex % N
# Driver code
arr = [ 3, -6, 7, -4, -4, 6, -1 ]
N = len(arr)
print(startingPoint(arr,N))
# This code is contributed by Virusbuddah
C
// C# program for the above approach
using System;
class GFG{
// Function to find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
static int startingPoint(int[] A, int N)
{
// Stores the sum of the array
int sum = 0;
// Stores the starting index
int ind = 0;
// Stores the minimum prefix
// sum of A[0..i]
int min = Int32.MaxValue;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Update the value of sum
sum += A[i];
// If sum is less than min
if (sum < min)
{
// Update the min as the
// value of prefix sum
min = sum;
// Update in
ind = i + 1;
}
}
// Otherwise, no such index is
// possible
if (sum < 0)
{
return -1;
}
return ind % N;
}
// Driver Code
public static void Main()
{
int[] arr = { 3, -6, 7, -4, -4, 6, -1 };
int N = arr.Length;
Console.Write(startingPoint(arr, N));
}
}
// This code is contributed by ukasp
java 描述语言
<script>
// javascript program for the above approach
// Function to find the starting index
// of the given circular array s.t.
// prefix sum array is non negative
function startingPoint(A , N) {
// Stores the sum of the array
var sum = 0;
// Stores the starting index
var it = 0;
// Stores the minimum prefix
// sum of A[0..i]
var min = Number.MAX_VALUE;
// Traverse the array arr
for (i = 0; i < N; i++) {
// Update the value of sum
sum += A[i];
// If sum is less than min
if (sum < min) {
// Update the min as the
// value of prefix sum
min = sum;
// Update in
it = i + 1;
}
}
// Otherwise, no such index is
// possible
if (sum < 0) {
return -1;
}
return it % N;
}
// Driver Code
var arr = [ 3, -6, 7, -4, -4, 6, -1 ];
var N = arr.length;
document.write(startingPoint(arr, N));
// This code contributed by umadevi9616
</script>
Output:
5
时间复杂度: O(N) 辅助空间: O(1)
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