Given two arrays arr1[] and arr2[] of size N and M (M < N) add all elements of arr2[] in all possible subarrays of arr1[] of size M exactly once. The task is to return the final array arr1[] after performing the above operations.
Examples:
Input: arr1[] = {1, 1, 1, 1, 1}, arr2[] = {1, 1, 1}
Output: 2 3 4 3 2
Explanation: There are exactly 3 subarrays of size M in arr1[]
adding corresponding elements of arr2[] in first subarray from element 1 to 3. arr1[] becomes {2, 2, 2, 1, 1}
adding corresponding elements of arr2[] in second subarray from element 2 to 4. arr1[] becomes {2, 3, 3, 2, 1}
adding corresponding elements of arr2[] in the third subarray from elements 3 to 5. arr1[] becomes {2, 3, 4, 3, 2}Input: arr1[] = {1, 2, 3, 4, 5}, arr2[] = {10}
Output: 11 12 13 14 15
Naive Approach: The problem can be solved using linear iteration based on the following idea:
Keep adding all elements of arr2[] in all possible subarrays of size M of arr1[]. There will be exactly N – M + 1 subarrays.
Follow the steps mentioned below to implement the idea:
- Iterate from i = 0 to N-M+1:
- Iterate from j = 0 to M:
- Add arr2[j] with arr[i+j.
- Iterate from j = 0 to M:
- Return the final state of arr1[] as the required answer.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to find the final state of arr1[]void addArr2ToArr1(int arr1[], int arr2[], int N, int M){ // Iterating through all possible subarrays // of size M in arr1[] there will be // exactly N - M + 1 subarrays for (int i = 0; i < N - M + 1; i++) { // Iterating through M elements of arr2[] // and adding in arr1[]'s 'i'th subarray's // 'i + j' th element for (int j = 0; j < M; j++) { arr1[i + j] = arr1[i + j] + arr2[j]; } } // Printing finally array for (int i = 0; i < N; i++) { cout << arr1[i] << " "; } cout << endl;}// Driver programint main(){ int arr1[] = { 1, 1, 1, 1, 1 }, arr2[] = { 1, 1, 1 }; int N = sizeof(arr1) / sizeof(arr1[0]); int M = sizeof(arr2) / sizeof(arr2[0]); // Function call for testcase 1 addArr2ToArr1(arr1, arr2, N, M); int arr3[] = { 10, 11, 12, 13, 14 }, arr4[] = { 10 }; int N2 = sizeof(arr3) / sizeof(arr3[0]); int M2 = sizeof(arr4) / sizeof(arr4[0]); // Function call for testcase 2 addArr2ToArr1(arr3, arr4, N2, M2); int arr5[] = { 12, 11, 10, 9 }, arr6[] = { 1, 2, 3, 4, 5 }; int N3 = sizeof(arr5) / sizeof(arr5[0]); int M3 = sizeof(arr6) / sizeof(arr6[0]); // Function call for testcase 3 addArr2ToArr1(arr5, arr6, N3, M3); return 0;} |
Java
// Java code to implement the approachimport java.io.*;class GFG { // Function to find the final state of arr1[] static void addArr2ToArr1(int[] arr1, int[] arr2, int N, int M) { // Iterating through all possible subarrays // of size M in arr1[] there will be // exactly N - M + 1 subarrays for (int i = 0; i < N - M + 1; i++) { // Iterating through M elements of arr2[] // and adding in arr1[]'s 'i'th subarray's // 'i + j' th element for (int j = 0; j < M; j++) { arr1[i + j] = arr1[i + j] + arr2[j]; } } // Printing finally array for (int i = 0; i < N; i++) { System.out.print(arr1[i] + " "); } System.out.println(); } public static void main(String[] args) { int[] arr1 = { 1, 1, 1, 1, 1 }; int[] arr2 = { 1, 1, 1 }; int N = arr1.length; int M = arr2.length; // Function call for testcase 1 addArr2ToArr1(arr1, arr2, N, M); int[] arr3 = { 10, 11, 12, 13, 14 }; int[] arr4 = { 10 }; int N2 = arr3.length; int M2 = arr4.length; // Function call for testcase 2 addArr2ToArr1(arr3, arr4, N2, M2); int[] arr5 = { 12, 11, 10, 9 }; int[] arr6 = { 1, 2, 3, 4, 5 }; int N3 = arr5.length; int M3 = arr6.length; // Function call for testcase 3 addArr2ToArr1(arr5, arr6, N3, M3); }}// This code is contributed by lokesh. |
Python3
# python code# Function to find the final state of arr1[]def addArr2ToArr1(arr1, arr2, N, M): # Iterating through all possible subarrays # of size M in arr1[] there will be # exactly N - M + 1 subarrays for i in range(0, N - M + 1): # Iterating through M elements of arr2[] # and adding in arr1[]'s 'i'th subarray's # 'i + j' th element for j in range(0, M): arr1[i + j] = arr1[i + j] + arr2[j] # Printing finally array for i in range(0, N): print(arr1[i], end=" ") print()# Driver programarr1 = [1, 1, 1, 1, 1]arr2 = [1, 1, 1]N = len(arr1)M = len(arr2)# Function call for testcase 1addArr2ToArr1(arr1, arr2, N, M)arr3 = [10, 11, 12, 13, 14]arr4 = [10]N2 = len(arr3)M2 = len(arr4)# Function call for testcase 2addArr2ToArr1(arr3, arr4, N2, M2)arr5 = [12, 11, 10, 9]arr6 = [1, 2, 3, 4, 5]N3 = len(arr5)M3 = len(arr6)# Function call for testcase 3addArr2ToArr1(arr5, arr6, N3, M3)# This code is contributed by ksam24000 |
C#
// C# code to implement the approachusing System;class GFG { // Function to find the final state of arr1[] static void addArr2ToArr1(int[] arr1, int[] arr2, int N, int M) { // Iterating through all possible subarrays // of size M in arr1[] there will be // exactly N - M + 1 subarrays for (int i = 0; i < N - M + 1; i++) { // Iterating through M elements of arr2[] // and adding in arr1[]'s 'i'th subarray's // 'i + j' th element for (int j = 0; j < M; j++) { arr1[i + j] = arr1[i + j] + arr2[j]; } } // Printing finally array for (int i = 0; i < N; i++) { Console.Write(arr1[i] + " "); } Console.WriteLine(); } // Driver program public static void Main() { int[] arr1 = { 1, 1, 1, 1, 1 }; int[] arr2 = { 1, 1, 1 }; int N = arr1.Length; int M = arr2.Length; // Function call for testcase 1 addArr2ToArr1(arr1, arr2, N, M); int[] arr3 = { 10, 11, 12, 13, 14 }; int[] arr4 = { 10 }; int N2 = arr3.Length; int M2 = arr4.Length; // Function call for testcase 2 addArr2ToArr1(arr3, arr4, N2, M2); int[] arr5 = { 12, 11, 10, 9 }; int[] arr6 = { 1, 2, 3, 4, 5 }; int N3 = arr5.Length; int M3 = arr6.Length; // Function call for testcase 3 addArr2ToArr1(arr5, arr6, N3, M3); }}// This code is contributed by Samim Hossain Mondal. |
Javascript
// Function to find the final state of arr1[]function addArr2ToArr1(arr1, arr2, N, M) { // Iterating through all possible subarrays // of size M in arr1[] there will be // exactly N - M + 1 subarrays for (let i = 0; i < N - M + 1; i++) { // Iterating through M elements of arr2[] // and adding in arr1[]'s 'i'th subarray's // 'i + j' th element for (let j = 0; j < M; j++) { arr1[i + j] = arr1[i + j] + arr2[j]; } } // Printing finally array for (let i = 0; i < N; i++) { console.log(arr1[i]); } console.log();}// Driver programlet arr1 = [1, 1, 1, 1, 1];let arr2 = [1, 1, 1];let N = arr1.length;let M = arr2.length;// Function call for testcase 1addArr2ToArr1(arr1, arr2, N, M);let arr3 = [10, 11, 12, 13, 14];let arr4 = [10];let N2 = arr3.length;let M2 = arr4.length;// Function call for testcase 2addArr2ToArr1(arr3, arr4, N2, M2);let arr5 = [12, 11, 10, 9];let arr6 = [1, 2, 3, 4, 5];let N3 = arr5.length;let M3 = arr6.length;// Function call for testcase 3addArr2ToArr1(arr5, arr6, N3, M3); |
Output
2 3 4 3 2 20 21 22 23 24 12 11 10 9
Time Complexity: O(N * M)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following idea:
This can be observed every element i of arr1[] will have some range of elements from arr2[] that gets added.
For example.
in case of arr1[] = {0, 0, 0, 0, 0}, arr2[] = {1, 2, 3}
element 1 will have 1 in its sum. Range of elements that got added from arr2[] is [0, 0].
element 2 will have 1 and 2 in its sum. Range of elements that got added from arr2[] is [0, 1].
element 3 will have 1, 2, and 3 in its sum. Range of elements that got added from arr2[] is [0, 2].
element 4 will have 2 and 3 in its sum. Range of elements that got added of arr2[] is [1, 2].
element 5 will have 3 in its sum. Range of elements that got added of arr2[] is [2, 2].Observation: every element ‘i’ will have sum of elements of arr2[] from max(0, M – N + i) to min(i, M-1).
Sum from max(1, M – N + i) to min(i, M-1) can be calculated using prefix sum in constant time.
Follow the steps below to solve the problem:
- Initializing prefix[] array which will be used to store the sum of all ranges of arr2[].
- Initializing L = max(1, M – N + i) and R = min(i, M).
- Traverse from 1 to N, Adding prefix[R] – prefix[L – 1] in arr[i – 1] .
- print the final array
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to add corresponding// elements of arr2[] in all possible// subarrays of size M of arr1[]void addArr2ToArr1(int arr1[], int arr2[], int N, int M){ // Prefix sum array will be used // to store sum of all ranges of arr2[] int prefix[M + 1] = { 0 }; // Taking prefix sum for (int i = 1; i <= M; i++) { prefix[i] = prefix[i - 1] + arr2[i - 1]; } for (int i = 1; i <= N; i++) { // Add element i of arr1[] with range of // elements of arr2[] from l to r int r = min(i, M), l = max(1, M - N + i); // Every element 'i' of arr1[] will // have sum of elements of arr2[] from // max(1, M - N + i) to min(i, M-1) which // can be calculate using prefix // sum in constant time arr1[i - 1] = arr1[i - 1] + prefix[r] - prefix[l - 1]; } for (int i = 0; i < N; i++) { cout << arr1[i] << " "; }}// Driver programint main(){ int arr1[] = { 1, 1, 1, 1, 1 }, arr2[] = { 1, 1, 1 }; int N = sizeof(arr1) / sizeof(arr1[0]); int M = sizeof(arr2) / sizeof(arr2[0]); // Function call addArr2ToArr1(arr1, arr2, N, M); return 0;} |
Java
// Java code to implement the approachimport java.util.*;public class GFG { // Function to add corresponding // elements of arr2[] in all possible // subarrays of size M of arr1[] static void addArr2ToArr1(int arr1[], int arr2[], int N, int M) { // Prefix sum array will be used // to store sum of all ranges of arr2[] int prefix[] = new int[M + 1]; for (int i = 0; i < M + 1; i++) { prefix[i] = 0; } // Taking prefix sum for (int i = 1; i <= M; i++) { prefix[i] = prefix[i - 1] + arr2[i - 1]; } for (int i = 1; i <= N; i++) { // Add element i of arr1[] with range of // elements of arr2[] from l to r int r = Math.min(i, M); int l = Math.max(1, M - N + i); // Every element 'i' of arr1[] will // have sum of elements of arr2[] from // max(1, M - N + i) to min(i, M-1) which // can be calculate using prefix // sum in constant time arr1[i - 1] = arr1[i - 1] + prefix[r] - prefix[l - 1]; } for (int i = 0; i < N; i++) { System.out.print(arr1[i] + " "); } } // Driver program public static void main(String args[]) { int arr1[] = { 1, 1, 1, 1, 1 }; int arr2[] = { 1, 1, 1 }; int N = arr1.length; int M = arr2.length; // Function call addArr2ToArr1(arr1, arr2, N, M); }}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python code to implement the approachdef addArr2toArr1(arr1, arr2, n, m): # Prefix sum array will be used # to store sum for all ranges of arr2[] prefix = [] for i in range(m+1): prefix.append(0) # Taking prefix sum for i in range(1, m+1): prefix[i] = prefix[i-1] + arr2[i-1] for i in range(1, n+1): # Add elements i of arr1 with range of # elements of arr2 from 1 to r r = int(min(i, m)) l = int(max(1, m-n+i)) # Every element 'i' of arr1 will # have sum of elements of arr2 from # max(1, m-n+1) to min(1, m-1) which # can be calculated using prefix # sum in constant time arr1[i-1] = arr1[i-1] + prefix[r] - prefix[l-1] for i in range(n): print(arr1[i], end=" ")arr1 = [1, 1, 1, 1, 1]arr2 = [1, 1, 1]# function calladdArr2toArr1(arr1, arr2, len(arr1), len(arr2)) |
C#
// C# code to implement the approachusing System;public class GFG { // Function to add corresponding // elements of arr2[] in all possible // subarrays of size M of arr1[] static void addArr2ToArr1(int[] arr1, int[] arr2, int N, int M) { // Prefix sum array will be used // to store sum of all ranges of arr2[] int[] prefix = new int[M + 1]; for (int i = 0; i < M + 1; i++) { prefix[i] = 0; } // Taking prefix sum for (int i = 1; i <= M; i++) { prefix[i] = prefix[i - 1] + arr2[i - 1]; } for (int i = 1; i <= N; i++) { // Add element i of arr1[] with range of // elements of arr2[] from l to r int r = Math.Min(i, M); int l = Math.Max(1, M - N + i); // Every element 'i' of arr1[] will // have sum of elements of arr2[] from // max(1, M - N + i) to min(i, M-1) which // can be calculate using prefix // sum in constant time arr1[i - 1] = arr1[i - 1] + prefix[r] - prefix[l - 1]; } for (int i = 0; i < N; i++) { Console.Write(arr1[i] + " "); } } static public void Main() { // Code int[] arr1 = { 1, 1, 1, 1, 1 }; int[] arr2 = { 1, 1, 1 }; int N = arr1.Length; int M = arr2.Length; // Function call addArr2ToArr1(arr1, arr2, N, M); }}// This code is contributed by lokeshmvs21. |
Javascript
function addArr2toArr1(arr1, arr2, n, m) { // Prefix sum array will be used // to store sum for all ranges of arr2[] var prefix = []; for (var i = 0; i <= m; i++) { prefix.push(0); } // Taking prefix sum for (var i = 1; i <= m; i++) { prefix[i] = prefix[i - 1] + arr2[i - 1]; } for (var i = 1; i <= n; i++) { // Add elements i of arr1 with range of // elements of arr2 from 1 to r var r = Math.min(i, m); var l = Math.max(1, m - n + i); // Every element 'i' of arr1 will // have sum of elements of arr2 from // max(1, m-n+1) to min(1, m-1) which // can be calculated using prefix // sum in constant time arr1[i - 1] = arr1[i - 1] + prefix[r] - prefix[l - 1]; } for (var i = 0; i < n; i++) { console.log(arr1[i]); }}var arr1 = [1, 1, 1, 1, 1];var arr2 = [1, 1, 1];// function calladdArr2toArr1(arr1, arr2, arr1.length, arr2.length); |
Output
2 3 4 3 2
Time Complexity: O(M + N)
Auxiliary Space: O(M)
Related Articles:
- Introduction to Arrays – Data Structure and Algorithm Tutorials
- Prefix Sum Array – Implementation and Applications in Competitive Programming
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