给定大小为N的整数数组arr [] ，并以{L，R，X}的形式进行Q查询，任务是从给定数组中找到L与R之间的最小索引，以使arr [i]！= X。如果数组中不存在这样的索引，则打印-1 。
例子：
Input: arr[] = {1, 2, 1, 1, 3, 5}, query[][] = {{0, 3, 1}, {1, 5, 2}, {2, 3, 1}}
Output: 1
2
-1
Explanation:
The first index from 0 to 3 which does not contain 1 is 1
The first index from 1 to 5 which does not contain 2 is 2
All the indices from 2 to 3 contain 1. Hence, the answer is -1.
Input: arr[] = { 1, 2, 3, 2, 1, 1, 3, 5}, query[][] = { { 1, 4, 2 }, { 4, 5, 1 }, { 2, 3, 1 }, { 4, 6, 1 }}
Output: 2
-1
2
6
天真的方法：
遍历每个查询的范围[L，R] ，并检查是否存在不包含X的索引。如果找到这样的索引，请打印该索引。否则，打印-1 。
时间复杂度： O(N * Q)
辅助空间： O(1)
高效方法：
通过为每个数组元素预先计算并存储与当前元素不同的下一个元素的索引，可以进一步优化上述方法，从而将每个查询的计算复杂度降低到O(1)。
请按照以下步骤解决问题：
- 创建一个辅助数组nextpos []来为每个数组元素存储与当前元素不同的下一个元素的索引。
- 现在要处理每个查询，首先检查索引L上的值是否不等于X。如果是这样，那么答案将是L。
- 否则，意味着arr [L] = X。在这种情况下，我们需要找到下一个索引，该索引的值不同于arr [L] 。可以从nextpos [L]获得。
- 如果nextpos [L]小于等于R ，则打印nexpos [L]。
- 如果以上条件均不满足，则答案将为-1 。
插图：
arr[] = {1, 2, 1, 1, 3, 5}, query[][] = {{0, 3, 1}, {1, 5, 2}, {2, 3, 1}}
For the given arr[], nextpos[] array will be {1, 2, 4, 4, 5, 6}
For 1st Query: L = 0, R = 3, X = 1
arr[0] = 1 = X
nextpos[0] = 1( < 3)
Hence, the answer is 1.
For the 2nd Query: L = 1, R = 5, X = 2
arr[1] = 2( = X)
nextpos[1] = 2( < 5)
Hence, the answer is 2.
For the 3rd Query: L = 2, R = 3, X = 1
arr[2] = 1( = X)
nextpos[2] = 4( > R)
Hence, the answer is -1
下面是上述方法的实现：
C++
// C++ Program to find the smallest
// index in the array in the range
// [L, R] which does not contain X
#include
using namespace std;
// Precompute the index of next
// different element in the array
// for every array element
void precompute(int nextpos[], int arr[],
int N)
{
// Default value
nextpos[N - 1] = N;
for (int i = N - 2; i >= 0; i--) {
// Compute nextpos[i] using
// nextpos[i+1]
if (arr[i] == arr[i + 1])
nextpos[i] = nextpos[i + 1];
else
nextpos[i] = i + 1;
}
}
// Function to return the smallest index
void findIndex(int query[][3], int arr[],
int N, int Q)
{
// nextpos[i] will store the next
// position p where arr[p]!=arr[i]
int nextpos[N];
precompute(nextpos, arr, N);
for (int i = 0; i < Q; i++) {
int l, r, x;
l = query[i][0];
r = query[i][1];
x = query[i][2];
int ans = -1;
// If X is not present at l
if (arr[l] != x)
ans = l;
// Otherwise
else {
// Find the index which
// stores a value different
// from X
int d = nextpos[l];
// If that index is within
// the range
if (d <= r)
ans = d;
}
cout << ans << "\n";
}
}
// Driver Code
int main()
{
int N, Q;
N = 6;
Q = 3;
int arr[] = { 1, 2, 1, 1, 3, 5 };
int query[Q][3] = { { 0, 3, 1 },
{ 1, 5, 2 },
{ 2, 3, 1 } };
findIndex(query, arr, N, Q);
return 0;
} Java
// Java program to find the smallest
// index in the array in the range
// [L, R] which does not contain X
class GFG{
// Precompute the index of next
// different element in the array
// for every array element
static void precompute(int nextpos[], int arr[],
int N)
{
// Default value
nextpos[N - 1] = N;
for(int i = N - 2; i >= 0; i--)
{
// Compute nextpos[i] using
// nextpos[i+1]
if (arr[i] == arr[i + 1])
nextpos[i] = nextpos[i + 1];
else
nextpos[i] = i + 1;
}
}
// Function to return the smallest index
static void findIndex(int query[][], int arr[],
int N, int Q)
{
// nextpos[i] will store the next
// position p where arr[p]!=arr[i]
int []nextpos = new int[N];
precompute(nextpos, arr, N);
for(int i = 0; i < Q; i++)
{
int l, r, x;
l = query[i][0];
r = query[i][1];
x = query[i][2];
int ans = -1;
// If X is not present at l
if (arr[l] != x)
ans = l;
// Otherwise
else
{
// Find the index which
// stores a value different
// from X
int d = nextpos[l];
// If that index is within
// the range
if (d <= r)
ans = d;
}
System.out.print(ans + "\n");
}
}
// Driver Code
public static void main(String[] args)
{
int N, Q;
N = 6;
Q = 3;
int arr[] = { 1, 2, 1, 1, 3, 5 };
int query[][] = { { 0, 3, 1 },
{ 1, 5, 2 },
{ 2, 3, 1 } };
findIndex(query, arr, N, Q);
}
}
// This code is contributed by Amit KatiyarPython3
# Python3 program to find the smallest
# index in the array in the range
# [L, R] which does not contain X
# Precompute the index of next
# different element in the array
# for every array element
def precompute(nextpos, arr, N):
# Default value
nextpos[N - 1] = N
for i in range(N - 2, -1, -1):
# Compute nextpos[i] using
# nextpos[i+1]
if arr[i] == arr[i + 1]:
nextpos[i] = nextpos[i + 1]
else:
nextpos[i] = i + 1
# Function to return the smallest index
def findIndex(query, arr, N, Q):
# nextpos[i] will store the next
# position p where arr[p]!=arr[i]
nextpos = [0] * N
precompute(nextpos, arr, N)
for i in range(Q):
l = query[i][0]
r = query[i][1]
x = query[i][2]
ans = -1
# If X is not present at l
if arr[l] != x:
ans = l
# Otherwise
else:
# Find the index which
# stores a value different
# from X
d = nextpos[l]
# If that index is within
# the range
if d <= r:
ans = d
print(ans)
# Driver code
N = 6
Q = 3
arr = [ 1, 2, 1, 1, 3, 5 ]
query = [ [ 0, 3, 1 ],
[ 1, 5, 2 ],
[ 2, 3, 1 ] ]
findIndex(query, arr, N, Q)
# This code is contributed by divyeshrabadiya07C#
// C# program to find the smallest
// index in the array in the range
// [L, R] which does not contain X
using System;
class GFG{
// Precompute the index of next
// different element in the array
// for every array element
static void precompute(int []nextpos,
int []arr, int N)
{
// Default value
nextpos[N - 1] = N;
for(int i = N - 2; i >= 0; i--)
{
// Compute nextpos[i] using
// nextpos[i+1]
if (arr[i] == arr[i + 1])
nextpos[i] = nextpos[i + 1];
else
nextpos[i] = i + 1;
}
}
// Function to return the smallest index
static void findIndex(int [,]query, int []arr,
int N, int Q)
{
// nextpos[i] will store the next
// position p where arr[p]!=arr[i]
int []nextpos = new int[N];
precompute(nextpos, arr, N);
for(int i = 0; i < Q; i++)
{
int l, r, x;
l = query[i, 0];
r = query[i, 1];
x = query[i, 2];
int ans = -1;
// If X is not present at l
if (arr[l] != x)
ans = l;
// Otherwise
else
{
// Find the index which
// stores a value different
// from X
int d = nextpos[l];
// If that index is within
// the range
if (d <= r)
ans = d;
}
Console.Write(ans + "\n");
}
}
// Driver Code
public static void Main(String[] args)
{
int N, Q;
N = 6;
Q = 3;
int []arr = { 1, 2, 1, 1, 3, 5 };
int [,]query = { { 0, 3, 1 },
{ 1, 5, 2 },
{ 2, 3, 1 } };
findIndex(query, arr, N, Q);
}
}
// This code is contributed by Amit Katiyar输出：
1
2
-1
时间复杂度： O(N)
辅助空间： O(N)