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.


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

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[1000000];
li A[200000];

void build(int node, int start, int end)
    if(start == end)
        tree[node] = A[start];
        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];
        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; 
    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
      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->array=(long *)malloc(q->capacity*sizeof(long));
    q->index=(int *)malloc(q->capacity*sizeof(int));
    for(int i=0;i<q->capacity;i++){
    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){

void deque(struct queue *q){

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(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;
    long *a=(long *)malloc(n*sizeof(long));
    int d;
        long max,min=INT_MAX;
        struct queue *q=create(n);
        //printf("%d %d ",q->rear,q->size);
            if(empty(q))    enque(q,a[j],j); 
                while(a[j]>=bottom_array(q)){   deque_back(q);  }
            //for(k=0;k<n;k++){ printf("%li ",q->array[k]);  }
            //printf("%d ",bottom_array(q));
            while(j-top_index(q)<d && j<n){
                if(top_array(q)<min)    min=top_array(q);
            if(j-top_index(q)==d && j<n){
                if(top_array(q)<min)    min=top_array(q);
        if(top_array(q)<min)    min=top_array(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]
    for i in range(d, n):
        if l[0] + d <= i:
        while l and a[l[-1]] < a[i]:
        m = min(m, a[l[0]])

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