# Queries with Fixed Length

### Problem Statement :

```Consider an -integer sequence, . We perform a query on  by using an integer, , to calculate the result of the following expression:

In other words, if we let , then you need to calculate .

Given  and  queries, return a list of answers to each query.

Example

The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is .

The second query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is .

Return .

Function Description

Complete the solve function below.

solve has the following parameter(s):

int arr[n]: an array of integers
int queries[q]: the lengths of subarrays to query
Returns

int[q]: the answers to each query
Input Format

The first line consists of two space-separated integers, n  and q.
The second line consists of n space-separated integers, the elements of arr.
Each of the q  subsequent lines contains a single integer denoting the value of d for that query.```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

#define li long int
#define INT_MAX 1000000007
li tree;
li A;

void build(int node, int start, int end)
{
if(start == end)
{
tree[node] = A[start];
}
else
{
int mid = (start + end) / 2;
build(2*node, start, mid);
build(2*node+1, mid+1, end);

tree[node] = max(tree[2*node] , tree[2*node+1]);
}
}

int query(int node, int start, int end, int l, int r)
{
if(r < start or end < l)
{
return 0;
}
if(l <= start and end <= r)
{
return tree[node];
}

int mid = (start + end) / 2;
int p1 = query(2*node, start, mid, l, r);
int p2 = query(2*node+1, mid+1, end, l, r);
return max(p1 , p2);
}

int main() {

li q,n,d;

cin >> n >> q;

for(int i=0;i<n;++i) cin >> A[i];

build(1,0,n-1);

while(q--)
{

cin >> d;

li min=INT_MAX;
for(int i=0;i<=n-d;++i)
{
li temp_max=query(1,0,n-1,i,i+d-1);

if(temp_max<min) min = temp_max;

}

cout<<min<<endl;
}

return 0;
}

In Java :

import java.util.Scanner;

public class Solution
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
int dataCount = scan.nextInt();
int queriesCount = scan.nextInt();
int[] data = new int[dataCount];
for (int i = 0; i < dataCount; ++i)
data[i] = scan.nextInt();
for (int i = 0; i < queriesCount; ++i)
System.out.println(findMinMax(data, scan.nextInt()));
}

private static int findMinMax(int[] data, int range)
{
if (range >= data.length)
return easyMax(data, 0, data.length);
int maxV = easyMax(data, data.length - range, data.length);
int minV = maxV;
for (int i = data.length - range - 1; i >= 0; --i)
{
if (data[i] == data[i+range]) //nothing changed
continue;
if (data[i] > maxV)
maxV = data[i];
else if (data[i + range] == maxV)
maxV = easyMax(data, i, i + range);
if (maxV < minV)
minV = maxV;
}
return minV;
}

private static int easyMax(int[] data, int fromInclusive, int toExclusive)
{
int pos = toExclusive;
int max = data[--pos];
while (--pos >= fromInclusive)
if (data[pos] > max)
max = data[pos];
return max;
}
}

In C :

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>

struct queue{
int front,rear,size;
unsigned capacity;
long *array;
int *index;
};

struct queue *create(unsigned capacity){
struct queue *q=(struct queue *)malloc(sizeof(struct queue));
q->front=0;
q->rear=capacity-1;
q->size=0;
q->capacity=capacity;
q->array=(long *)malloc(q->capacity*sizeof(long));
q->index=(int *)malloc(q->capacity*sizeof(int));
for(int i=0;i<q->capacity;i++){
q->array[i]=0;
q->index[i]=0;
}
return q;
}

int full(struct queue* q){
if(q->size==q->capacity)    return 1;
else    return 0;
}

int empty(struct queue* q){
if(q->size==0)  return 1;
else    return 0;
}

void enque(struct queue* q, long x, int i){
if(!full(q)){
q->size++;
q->rear=(q->rear+1)%(q->capacity);
q->array[q->rear]=x;
q->index[q->rear]=i;
}
}

void deque(struct queue *q){
if(!empty(q)){
q->size--;
q->front=(q->front+1)%(q->capacity);
}
}

long top_array(struct queue *q){
if(!empty(q))   return q->array[q->front];
return INT_MAX;
}

int top_index(struct queue *q){
if(!empty(q))   return q->index[q->front];
return INT_MAX;
}

void deque_back(struct queue *q){
if(!empty(q)){
q->size--;
q->rear=(q->rear-1);
if(q->rear<0)   q->rear=q->capacity+q->rear;
}
}

long bottom_array(struct queue *q){
if(!empty(q))   return q->array[q->rear];
return INT_MAX;
}

int main(){

int n,Q,i,j,k;
scanf("%d%d",&n,&Q);
long *a=(long *)malloc(n*sizeof(long));
for(i=0;i<n;i++){
scanf("%li",&a[i]);
}
int d;

for(i=0;i<Q;i++){
long max,min=INT_MAX;
scanf("%d",&d);
struct queue *q=create(n);
j=0;
//printf("%d %d ",q->rear,q->size);
while(j<d){
if(empty(q))    enque(q,a[j],j);
else{
while(a[j]>=bottom_array(q)){   deque_back(q);  }
enque(q,a[j],j);
}
//for(k=0;k<n;k++){ printf("%li ",q->array[k]);  }
//printf("%d ",bottom_array(q));
j++;
}
min=top_array(q);

while(j<n){
while(j-top_index(q)<d && j<n){
if(top_array(q)<min)    min=top_array(q);
while(a[j]>=bottom_array(q)){
deque_back(q);
}
enque(q,a[j],j);
j++;
}
if(j-top_index(q)==d && j<n){
if(top_array(q)<min)    min=top_array(q);
deque(q);
while(a[j]>=bottom_array(q)){
deque_back(q);
}
enque(q,a[j],j);
j++;
}
}

if(top_array(q)<min)    min=top_array(q);
printf("%li\n",min);
free(q);
}

return 0;
}

In Python3 :

from collections import deque
n, q = map(int, input().split())
a = list(map(int, input().split()))
for _ in range(q):
d = int(input())
l = deque()
m = 0
for i in range(d - 1, -1, -1):
if m < a[i]:
m = a[i]
l.appendleft(i)
for i in range(d, n):
if l + d <= i:
l.popleft()
while l and a[l[-1]] < a[i]:
l.pop()
l.append(i)
m = min(m, a[l])
print(m)```
```

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