给定一个大小为N的数组arr[]和一个由M 个整数组成的数组Q[] ，每个都代表一个查询，每个查询Q[i]的任务是找到值大于或的数组元素的最小索引等于Q[i] 。如果不存在这样的索引，则打印-1 。
例子：
Input: arr[] = { 1, 9 }, Q[] = { 7, 10, 0 }
Output: 1 -1 0
Explanation:
The smallest index of arr[] whose value is greater than Q[0] is 1.
No such index exists in arr[] whose value is greater than Q[1].
The smallest index of arr[] whose value is greater than Q[2] is 0.
Therefore, the required output is 1 -1 0.
Input: arr[] = {2, 3, 4}, Q[] = {2, 3, 4}
Output: 0 1 2
方法：该问题可以使用二分搜索和前缀和技术解决。请按照以下步骤解决问题：
- 初始化一个数组，比如storeArrIdx[] ，形式为{ first, second }以存储数组元素和索引。
- 按数组元素的升序对数组storeArridx[]进行排序。
- 按升序对数组 arr[] 进行排序。
- 初始化一个数组，比如minIdx[] ，这样minIdx[i]存储其值大于或等于arr[i]的数组元素的最小索引。
- 使用变量i反向遍历数组storeArrIdx[] 。对于每个第i个索引，更新minIdx[i] = min(minIdx[i + 1], storeArrIdx[i].second) 。
- 使用变量i遍历数组Q[] 。对于每个第i个查询，在arr[] 中找到Q[i]的lower_bound 的索引，并检查获得的索引是否小于N。如果发现为真，则打印minIdx[i] 。
- 否则，打印-1 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find the smallest index
// of an array element whose value is
// less than or equal to Q[i]
void minimumIndex(vector& arr,
vector& Q)
{
// Stores size of array
int N = arr.size();
// Stores coun of queries
int M = Q.size();
// Store array elements along
// with the index
vector > storeArrIdx;
// Store smallest index of an array
// element whose value is greater
// than or equal to i
vector minIdx(N);
// Traverse the array
for (int i = 0; i < N; ++i) {
// Insert {arr[i], i} into
// storeArrIdx[]
storeArrIdx.push_back({ arr[i], i });
}
// Sort the array
sort(arr.begin(), arr.end());
// Sort the storeArrIdx
sort(storeArrIdx.begin(),
storeArrIdx.end());
// Stores index of arr[N - 1] in
// sorted order
minIdx[N - 1]
= storeArrIdx[N - 1].second;
// Traverse the array storeArrIdx[]
for (int i = N - 2; i >= 0; i--) {
// Update minIdx[i]
minIdx[i] = min(minIdx[i + 1],
storeArrIdx[i].second);
}
// Traverse the array Q[]
for (int i = 0; i < M; i++) {
// Store the index of
// lower_bound of Q[i]
int pos
= lower_bound(arr.begin(),
arr.end(), Q[i])
- arr.begin();
// If no index found whose value
// greater than or equal to arr[i]
if (pos == N) {
cout << -1 << " ";
continue;
}
// Print smallest index whose value
// greater than or equal to Q[i]
cout << minIdx[pos] << " ";
}
}
// Driver Code
int main()
{
vector arr = { 1, 9 };
vector Q = { 7, 10, 0 };
minimumIndex(arr, Q);
return 0;
} Java
// Java program for above approach
import java.util.*;
import java.lang.*;
class pair
{
int element,index;
pair(int element, int index)
{
this.element = element;
this.index = index;
}
}
class GFG
{
// Function to find the smallest index
// of an array element whose value is
// less than or equal to Q[i]
static void minimumIndex(int[] arr,
int[] Q)
{
// Stores size of array
int N = arr.length;
// Stores coun of queries
int M = Q.length;
// Store array elements along
// with the index
ArrayList storeArrIdx = new ArrayList<>();
// Store smallest index of an array
// element whose value is greater
// than or equal to i
int[] minIdx = new int[N];
// Traverse the array
for (int i = 0; i < N; ++i)
{
// Insert {arr[i], i} into
// storeArrIdx[]
storeArrIdx.add(new pair(arr[i], i));
}
// Sort the array
Arrays.sort(arr);
// Sort the storeArrIdx
Collections.sort(storeArrIdx, (a, b)->a.element-b.element);
// Stores index of arr[N - 1] in
// sorted order
minIdx[N - 1]
= storeArrIdx.get(N - 1).index;
// Traverse the array storeArrIdx[]
for (int i = N - 2; i >= 0; i--) {
// Update minIdx[i]
minIdx[i] =Math.min(minIdx[i + 1],
storeArrIdx.get(i).index);
}
// Traverse the array Q[]
for (int i = 0; i < M; i++) {
// Store the index of
// lower_bound of Q[i]
int pos
= lower_bound(arr, Q[i]);
// If no index found whose value
// greater than or equal to arr[i]
if (pos == N) {
System.out.print("-1"+" ");
continue;
}
// Print smallest index whose value
// greater than or equal to Q[i]
System.out.print(minIdx[pos]+" ");
}
}
static int lower_bound(int[] arr,int element)
{
for(int i = 0; i < arr.length; i++)
if(element <= arr[i])
return i;
return arr.length;
}
// Driver function
public static void main (String[] args)
{
int[] arr = { 1, 9 };
int[] Q = { 7, 10, 0 };
minimumIndex(arr, Q);
}
}
// This code is contributed by offbeat Python3
# Python3 program to implement
# the above approachf
from bisect import bisect_left
# Function to find the smallest index
# of an array element whose value is
# less than or equal to Q[i]
def minimumIndex(arr, Q):
# Stores size of array
N = len(arr)
# Stores coun of queries
M = len(Q)
# Store array elements along
# with the index
storeArrIdx = []
# Store smallest index of an array
# element whose value is greater
# than or equal to i
minIdx = [0]*(N)
# Traverse the array
for i in range(N):
# Insert {arr[i], i} into
# storeArrIdx[]
storeArrIdx.append([arr[i], i])
# Sort the array
arr = sorted(arr)
# Sort the storeArrIdx
storeArrIdx = sorted(storeArrIdx)
# Stores index of arr[N - 1] in
# sorted order
minIdx[N - 1] = storeArrIdx[N - 1][1]
# Traverse the array storeArrIdx[]
for i in range(N - 2, -1, -1):
# Update minIdx[i]
minIdx[i] = min(minIdx[i + 1], storeArrIdx[i][1])
# Traverse the array Q[]
for i in range(M):
# Store the index of
# lower_bound of Q[i]
pos = bisect_left(arr, Q[i])
# If no index found whose value
# greater than or equal to arr[i]
if (pos == N):
print(-1, end = " ")
continue
# Prsmallest index whose value
# greater than or equal to Q[i]
print(minIdx[pos], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [1, 9]
Q = [7, 10, 0]
minimumIndex(arr, Q)
# This code is contributed by mohit kumar 29输出：
1 -1 0时间复杂度： O(N * log(N))
辅助空间： O(N)
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