Given an array arr[] of size N and an element K. The task is to find the number of sub-arrays of the given array such that the remainder when dividing the sum of its elements by K is equal to the number of elements in the subarray.
Examples:
Input: arr[] = {1, 4, 2, 3, 5}, K = 4
Output: 4
{1}, {1, 4, 2}, {4, 2} and {5}
are the only valid subarrays.
Input: arr[] = {4, 2, 4, 2, 4, 2, 4, 2}, K = 4
Output: 7
Approach: Let’s define a sequence Sn such that Si = A1 + A2 + ··· + Ai and S0 = 0. Then, the condition that a contiguous subsequence Ai+1, …, Aj is valid can be represented as (Sj – Si) % K = j – i.
This equation can then be transformed into the following equivalent conditions:
(Sj – j) % K = (Si – i) % K and j – i < K.
Therefore, for each j(1 ? j ? N), count the number of j – K < i < j such that (Sj – j) % K = (Si – i) % K. For j the segment needed to be searched is (j – K, j), and for j + 1, it is (j – K + 1, j + 1), and these differ only by one element at the leftmost and rightmost, so in order to search for (j + 1)th after searching for jth element, only discard the leftmost element and add the rightmost element. Operations of discarding or adding can be performed quickly by managing the number of Si – i‘s by using associative arrays (such as map in C++ or dict in Python).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the number of subarrays// of the given array such that the remainder// when dividing the sum of its elements// by K is equal to the number of its elementsint sub_arrays(int a[], int n, int k){ // To store prefix sum int sum[n + 2] = { 0 }; for (int i = 0; i < n; i++) { // We are dealing with zero // indexed array a[i]--; // Taking modulus value a[i] %= k; // Prefix sum sum[i + 1] += sum[i] + a[i]; sum[i + 1] %= k; } // To store the required answer, the left // index and the right index int ans = 0, l = 0, r = 0; // To store si - i value map<int, int> mp; for (int i = 0; i < n + 1; i++) { // Include sum ans += mp[sum[i]]; mp[sum[i]]++; // Increment the right index r++; // If subarray has at least // k elements if (r - l >= k) { mp[sum[l]]--; l++; } } // Return the required answer return ans;}// Driver codeint main(){ int a[] = { 1, 4, 2, 3, 5 }; int n = sizeof(a) / sizeof(a[0]); int k = 4; // Function call cout << sub_arrays(a, n, k); return 0;} |
Java
// Java implementation of the approach import java.util.*; class gfg{ // Function to return the number of subarrays // of the given array such that the remainder // when dividing the sum of its elements // by K is equal to the number of its elements static int sub_arrays(int []a, int n, int k) { // To store prefix sum int sum[] = new int[n + 2] ; for (int i = 0; i < n+2; i++) { sum[i] = 0; } for (int i = 0; i < n; i++) { // We are dealing with zero // indexed array a[i]--; // Taking modulus value a[i] %= k; // Prefix sum sum[i + 1] += sum[i] + a[i]; sum[i + 1] %= k; } // To store the required answer, the left // index and the right index int ans = 0, l = 0, r = 0; // To store si - i value HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); for (int i = 0; i < n + 1; i++) { mp.put(sum[i], 0); } int temp; for (int i = 0; i < n + 1; i++) { // Include sum ans += (int)mp.get(sum[i]); temp =(int)mp.get(sum[i]) + 1; mp.put(sum[i], temp); // Increment the right index r++; // If subarray has at least // k elements if (r - l >= k) { //mp[sum[l]]--; temp = (int)mp.get(sum[l]) - 1; mp.put(sum[l], temp); l++; } } // Return the required answer return ans; } // Driver code public static void main(String args[]) { int []a = { 1, 4, 2, 3, 5 }; int n = a.length; int k = 4; // Function call System.out.print(sub_arrays(a, n, k)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach# Function to return the number of # subarrays of the given array # such that the remainder when dividing # the sum of its elements by K is # equal to the number of its elementsdef sub_arrays(a, n, k): # To store prefix sum sum = [0 for i in range(n + 2)] for i in range(n): # We are dealing with zero # indexed array a[i] -= 1 # Taking modulus value a[i] %= k # Prefix sum sum[i + 1] += sum[i] + a[i] sum[i + 1] %= k # To store the required answer, # the left index and the right index ans = 0 l = 0 r = 0 # To store si - i value mp = dict() for i in range(n + 1): # Include sum if sum[i] in mp: ans += mp[sum[i]] mp[sum[i]] = mp.get(sum[i], 0) + 1 # Increment the right index r += 1 # If subarray has at least # k elements if (r - l >= k): mp[sum[l]] -= 1 l += 1 # Return the required answer return ans# Driver codea = [1, 4, 2, 3, 5]n = len(a)k = 4# Function callprint(sub_arrays(a, n, k))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approach using System;using System.Collections.Generic;class gfg{ // Function to return the number of subarrays // of the given array such that the remainder // when dividing the sum of its elements // by K is equal to the number of its elements static int sub_arrays(int []a, int n, int k) { // To store prefix sum int []sum = new int[n + 2] ; for (int i = 0; i < n + 2; i++) { sum[i] = 0; } for (int i = 0; i < n; i++) { // We are dealing with zero // indexed array a[i]--; // Taking modulus value a[i] %= k; // Prefix sum sum[i + 1] += sum[i] + a[i]; sum[i + 1] %= k; } // To store the required answer, the left // index and the right index int ans = 0, l = 0, r = 0; // To store si - i value Dictionary<int, int> mp = new Dictionary<int, int>(); for (int i = 0; i < n + 1; i++) { if(!mp.ContainsKey(sum[i])) mp.Add(sum[i], 0); } int temp; for (int i = 0; i < n + 1; i++) { // Include sum ans += (int)mp[sum[i]]; temp =(int)mp[sum[i]] + 1; mp[sum[i]] = temp; // Increment the right index r++; // If subarray has at least // k elements if (r - l >= k) { //mp[sum[l]]--; temp = (int)mp[sum[l]] - 1; mp[sum[i]] = temp; l++; } } // Return the required answer return ans; } // Driver code public static void Main(String []args) { int []a = { 1, 4, 2, 3, 5 }; int n = a.Length; int k = 4; // Function call Console.Write(sub_arrays(a, n, k)); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation of the approach// Function to return the number of subarrays// of the given array such that the remainder// when dividing the sum of its elements// by K is equal to the number of its elementsfunction sub_arrays(a, n, k) { // To store prefix sum let sum = new Array(n + 2); for (let i = 0; i < n + 2; i++) { sum[i] = 0; } for (let i = 0; i < n; i++) { // We are dealing with zero // indexed array a[i]--; // Taking modulus value a[i] %= k; // Prefix sum sum[i + 1] += sum[i] + a[i]; sum[i + 1] %= k; } // To store the required answer, the left // index and the right index let ans = 0, l = 0, r = 0; // To store si - i value let mp = new Map(); for (let i = 0; i < n + 1; i++) { if (!mp.has(sum[i])) mp.set(sum[i], 0); } let temp; for (let i = 0; i < n + 1; i++) { // Include sum ans += mp.get(sum[i]); temp = mp.get(sum[i]) + 1; mp.set(sum[i], temp); // Increment the right index r++; // If subarray has at least // k elements if (r - l >= k) { //mp[sum[l]]--; temp = mp.get(sum[l]) - 1; mp.set(sum[i], temp); l++; } } // Return the required answer return ans;}// Driver codelet a = [1, 4, 2, 3, 5];let n = a.length;let k = 4;// Function calldocument.write(sub_arrays(a, n, k));// This code is contributed by _saurabh_jaiswal</script> |
Output:
4
Time Complexity: O(N* log(N))
Auxiliary Space: O(N)
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