Given an array arr[] consisting of N integers and an array Q[][2] consisting of M queries of the form {L, R}, the task for each query is to check if array elements over the range [L, R] forms an Arithmetic Progression or not. If found to be true, print “Yes”. Otherwise, print “No”.
Examples:
Input: arr[] = {1, 3, 5, 7, 6, 5, 4, 1}, Q[][] = {{0, 3}, {3, 4}, {2, 4}}
Output:
Yes
Yes
No
Explanation:
Query 1: The elements of the array over the range [0, 3] are {1, 3, 5, 7} which forms an arithmetic series with a common difference of 2. Hence, print “Yes”.
Query 2: The elements of the array over the range [3, 4 are {7, 6} which forms an arithmetic series with a common difference of -1. Hence, print “Yes”.
Query 3: The elements of the array over the range [2, 4 are {5, 7, 6}, which does not form an arithmetic series. Hence, print “Yes”.Input: arr[] = {1, 2}, Q[][] = {{0, 0}, {0, 1}, {0, 1}}
Output:
Yes
Yes
Yes
Naive Approach: The simplest approach to solve the problem is to traverse the given array over the range [L, R] for each query and check if the common difference between all the adjacent elements is the same or not. If the difference is the same, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if the given range// of queries form an AP or not in the// given array arr[]void findAPSequence(int arr[], int N, int Q[][2], int M){ // Traversing for all the Queries for (int i = 0; i < M; i++) { if (Q[i][0] == Q[i][1]) { cout << "Yes" << endl; } else { int ans = arr[Q[i][0] + 1] - arr[Q[i][0]]; bool flag = true; for (int j = Q[i][0]; j < Q[i][1]; j++) { // Check if difference is same or not. // If same then continue otherwise // print no. if (ans != (arr[j + 1] - arr[j])) { cout << "No" << endl; flag = false; break; } } // If all differences in diven range // are same then print yes if (flag) { cout << "Yes" << endl; } } }}// Driver Codeint main(){ int arr[] = { 1, 3, 5, 7, 6, 5, 4, 1 }; int Q[][2] = { { 0, 3 }, { 3, 4 }, { 2, 4 } }; int N = sizeof(arr) / sizeof(arr[0]); int M = sizeof(Q) / sizeof(Q[0]); findAPSequence(arr, N, Q, M); return 0;}// This code is contributed by Pushpesh Raj. |
Java
import java.util.Arrays;class GFG { static void findAPSequence(int[] arr, int N, int[][] Q, int M) { // Traversing for all the Queries for (int i = 0; i < M; i++) { if (Q[i][0] == Q[i][1]) { System.out.println("Yes"); } else { int ans = arr[Q[i][0] + 1] - arr[Q[i][0]]; boolean flag = true; for (int j = Q[i][0]; j < Q[i][1]; j++) { // Check if difference is same or not. // If same then continue otherwise // print no. if (ans != (arr[j + 1] - arr[j])) { System.out.println("No"); flag = false; break; } } // If all differences in diven range // are same then print yes if (flag) { System.out.println("Yes"); } } } } public static void main(String[] args) { int[] arr = { 1, 3, 5, 7, 6, 5, 4, 1 }; int[][] Q = { { 0, 3 }, { 3, 4 }, { 2, 4 } }; int N = arr.length; int M = Q.length; findAPSequence(arr, N, Q, M); }}// This code is contributed by aadityaburujwale. |
Python3
def findAPSequence(arr, N, Q, M): # Traversing for all the Queries for i in range(M): if Q[i][0] == Q[i][1]: print("Yes") else: ans = arr[Q[i][0] + 1] - arr[Q[i][0]] flag = True for j in range(Q[i][0], Q[i][1]): # Check if difference is same or not. # If same then continue otherwise # print no. if ans != (arr[j + 1] - arr[j]): print("No") flag = False break # If all differences in diven range # are same then print yes if flag: print("Yes")# Driver codearr = [1, 3, 5, 7, 6, 5, 4, 1]Q = [[0, 3], [3, 4], [2, 4]]N = len(arr)M = len(Q)findAPSequence(arr, N, Q, M)# This code is contributed by aadityaburujwale. |
C#
using System;using System.Linq;class GFG{ static void findAPSequence(int[] arr, int N, int[,] Q, int M) { // Traversing for all the Queries for (int i = 0; i < M; i++) { if (Q[i, 0] == Q[i, 1]) { Console.WriteLine("Yes"); } else { int ans = arr[Q[i, 0] + 1] - arr[Q[i, 0]]; bool flag = true; for (int j = Q[i, 0]; j < Q[i, 1]; j++) { // Check if difference is same or not. // If same then continue otherwise // print no. if (ans != (arr[j + 1] - arr[j])) { Console.WriteLine("No"); flag = false; break; } } // If all differences in diven range // are same then print yes if (flag) { Console.WriteLine("Yes"); } } } } // Driver code public static void Main(string[] args) { int[] arr = { 1, 3, 5, 7, 6, 5, 4, 1 }; int[,] Q = { { 0, 3 }, { 3, 4 }, { 2, 4 } }; int N = arr.Length; int M = Q.GetLength(0); // Function call findAPSequence(arr, N, Q, M); }}// This code is contributed by phasing17. |
Javascript
// Javascript program for the above approachfunction findAPSequence(arr, N, Q, M) { // Traversing for all the Queries for (let i = 0; i < M; i++) { if (Q[i][0] === Q[i][1]) { console.log("Yes"); } else { let ans = arr[Q[i][0] + 1] - arr[Q[i][0]]; let flag = true; for (let j = Q[i][0]; j < Q[i][1]; j++) { // Check if difference is same or not. // If same then continue otherwise // print no. if (ans !== (arr[j + 1] - arr[j])) { console.log("No"); flag = false; break; } } // If all differences in diven range // are same then print yes if (flag) { console.log("Yes"); } } }}// Driver codelet arr = [1, 3, 5, 7, 6, 5, 4, 1];let Q = [[0, 3], [3, 4], [2, 4]];let N = arr.length;let M = Q.length;findAPSequence(arr, N, Q, M);// This code is contributed by lokeshpotta20. |
Output
Yes Yes No
Time Complexity: O(N*M)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
- The idea is to precompute the longest length of the subarray forming an AP starting from any index i for every ith element of the array in an auxiliary array say dp[] using the Two Pointer Algorithm.
- For a given range [L, R], if the value of dp[L] is greater than or equal to (R – L), then the range will always form an AP as (R – L) is the current range of elements and dp[L] stores the length of the longest subarray forming AP from index L, then the subarray length must be smaller than dp[L].
Follow the steps below to solve the problem:
- Initialize an array say dp[] to store the length of the longest subarray starting from each index for each element at that index.
- Iterate over the range [0, N] using the variable i and perform the following steps:
- Initialize a variable say j as (i + 1) to store the last index array forming Arithmetic Progression from index i.
- Increment the value of j until (j + 1 < N) and (arr[j] – arr[j – 1]) is same as (arr[i + 1] – arr[i]).
- Iterate over the range [i, j – 1] using the variable, say K, and update the value of dp[K] as (j – K).
- Update the value of i as j.
- Traverse the given array of queries Q[] and for each query {L, R} if the value of dp[L] is greater than or equal to (R – L), then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if the given range// of queries form an AP or not in the// given array arr[]void findAPSequence(int arr[], int N, int Q[][2], int M){ // Stores length of the longest // subarray forming AP for every // array element int dp[N + 5] = { 0 }; // Iterate over the range [0, N] for (int i = 0; i + 1 < N;) { // Stores the index of the last // element of forming AP int j = i + 1; // Iterate until the element at // index (j, j + 1) forms AP while (j + 1 < N && arr[j + 1] - arr[j] == arr[i + 1] - arr[i]) // Increment j by 1 j++; // Traverse the current subarray // over the range [i, j - 1] for (int k = i; k < j; k++) { // Update the length of the // longest subarray at index k dp[k] = j - k; } // Update the value of i i = j; } // Traverse the given queries for (int i = 0; i < M; i++) { // Print the result if (dp[Q[i][0]] >= Q[i][1] - Q[i][0]) { cout << "Yes" << endl; } // Otherwise else { cout << "No" << endl; } }}// Driver Codeint main(){ int arr[] = { 1, 3, 5, 7, 6, 5, 4, 1 }; int Q[][2] = { { 0, 3 }, { 3, 4 }, { 2, 4 } }; int N = sizeof(arr) / sizeof(arr[0]); int M = sizeof(Q) / sizeof(Q[0]); findAPSequence(arr, N, Q, M); return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG{// Function to check if the given range// of queries form an AP or not in the// given array arr[]static void findAPSequence(int arr[], int N, int Q[][], int M){ // Stores length of the longest // subarray forming AP for every // array element int dp[] = new int[N + 5]; // Iterate over the range [0, N] for(int i = 0; i + 1 < N;) { // Stores the index of the last // element of forming AP int j = i + 1; // Iterate until the element at // index (j, j + 1) forms AP while (j + 1 < N && arr[j + 1] - arr[j] == arr[i + 1] - arr[i]) // Increment j by 1 j++; // Traverse the current subarray // over the range [i, j - 1] for(int k = i; k < j; k++) { // Update the length of the // longest subarray at index k dp[k] = j - k; } // Update the value of i i = j; } // Traverse the given queries for(int i = 0; i < M; i++) { // Print the result if (dp[Q[i][0]] >= Q[i][1] - Q[i][0]) { System.out.println("Yes"); } // Otherwise else { System.out.println("No"); } }}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 3, 5, 7, 6, 5, 4, 1 }; int Q[][] = { { 0, 3 }, { 3, 4 }, { 2, 4 } }; int N = arr.length; int M = Q.length; findAPSequence(arr, N, Q, M);}}// This code is contributed by Kingash |
Python3
# Python3 program for the above approach# Function to check if the given range# of queries form an AP or not in the# given array arr[]def findAPSequence(arr, N, Q, M): # Stores length of the longest # subarray forming AP for every # array element dp = [0] * (N + 5) # Iterate over the range [0, N] i = 0 while i + 1 < N: # Stores the index of the last # element of forming AP j = i + 1 # Iterate until the element at # index (j, j + 1) forms AP while (j + 1 < N and arr[j + 1] - arr[j] == arr[i + 1] - arr[i]): # Increment j by 1 j += 1 # Traverse the current subarray # over the range [i, j - 1] for k in range(i, j): # Update the length of the # longest subarray at index k dp[k] = j - k # Update the value of i i = j # Traverse the given queries for i in range(M): # Print the result if (dp[Q[i][0]] >= Q[i][1] - Q[i][0]): print("Yes") # Otherwise else: print("No")# Driver Codeif __name__ == "__main__": arr = [ 1, 3, 5, 7, 6, 5, 4, 1 ] Q = [ [ 0, 3 ], [ 3, 4 ], [ 2, 4 ] ] N = len(arr) M = len(Q) findAPSequence(arr, N, Q, M)# This code is contributed by ukasp |
C#
// C# code for above approachusing System;public class GFG{ // Function to check if the given range// of queries form an AP or not in the// given array arr[]static void findAPSequence(int[] arr, int N, int[, ] Q, int M){ // Stores length of the longest // subarray forming AP for every // array element int[] dp = new int[N + 5]; // Iterate over the range [0, N] for(int i = 0; i + 1 < N;) { // Stores the index of the last // element of forming AP int j = i + 1; // Iterate until the element at // index (j, j + 1) forms AP while (j + 1 < N && arr[j + 1] - arr[j] == arr[i + 1] - arr[i]) // Increment j by 1 j++; // Traverse the current subarray // over the range [i, j - 1] for(int k = i; k < j; k++) { // Update the length of the // longest subarray at index k dp[k] = j - k; } // Update the value of i i = j; } // Traverse the given queries for(int i = 0; i < M; i++) { // Print the result if (dp[Q[i, 0]] >= Q[i, 1] - Q[i, 0]) { Console.WriteLine("Yes"); } // Otherwise else { Console.WriteLine("No"); } }} static public void Main (){ int[] arr = { 1, 3, 5, 7, 6, 5, 4, 1 }; int[, ] Q = { { 0, 3 }, { 3, 4 }, { 2, 4 } }; int N = arr.Length; int M = Q.GetLength(0); findAPSequence(arr, N, Q, M); }}// This code is contributed by offbeat |
Javascript
<script>// javascript program for the above approach // Function to check if the given range // of queries form an AP or not in the // given array arr function findAPSequence(arr , N , Q , M) { // Stores length of the longest // subarray forming AP for every // array element var dp = Array(N + 5).fill(0); // Iterate over the range [0, N] for (i = 0; i + 1 < N;) { // Stores the index of the last // element of forming AP var j = i + 1; // Iterate until the element at // index (j, j + 1) forms AP while (j + 1 < N && arr[j + 1] - arr[j] == arr[i + 1] - arr[i]) // Increment j by 1 j++; // Traverse the current subarray // over the range [i, j - 1] for (k = i; k < j; k++) { // Update the length of the // longest subarray at index k dp[k] = j - k; } // Update the value of i i = j; } // Traverse the given queries for (i = 0; i < M; i++) { // Print the result if (dp[Q[i][0]] >= Q[i][1] - Q[i][0]) { document.write("Yes <br/>"); } // Otherwise else { document.write("No <br/>"); } } } // Driver Code var arr = [ 1, 3, 5, 7, 6, 5, 4, 1 ]; var Q = [ [ 0, 3 ], [ 3, 4 ], [ 2, 4 ] ]; var N = arr.length; var M = Q.length; findAPSequence(arr, N, Q, M);// This code contributed by umadevi9616</script> |
Output:
Yes Yes No
Time Complexity: O(N + M)
Auxiliary Space: O(N)
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