Print modified array after multiple array range increment operations
Given an array containing n integers and a value d. m queries are given. Each query has two values start and end. For each query, the problem is to increment the values from the start to end index in the given array by the given value d. A linear time-efficient solution is required for handling such multiple queries.
Examples:
Input : arr[] = {3, 5, 4, 8, 6, 1} Query list: {0, 3}, {4, 5}, {1, 4}, {0, 1}, {2, 5} d = 2 Output : 7 11 10 14 12 5 Executing 1st query {0, 3} arr = {5, 7, 6, 10, 6, 1} Executing 2nd query {4, 5} arr = {5, 7, 6, 10, 8, 3} Executing 3rd query {1, 4} arr = {5, 9, 8, 12, 10, 3} Executing 4th query {0, 1} arr = {7, 11, 8, 12, 10, 3} Executing 5th query {2, 5} arr = {7, 11, 10, 14, 12, 5} Note: Each query is executed on the previously modified array.
Naive Approach: For each query, traverse the array in the range start to end and increment the values in that range by the given value d.
Efficient Approach: Create an array sum[] of size n and initialize all of its index with the value 0. Now for each (start, end) index pair apply the given operation on the sum[] array. The operations are: sum[start] += d and sum[end+1] -= d only if the index (end+1) exists. Now, from the index i = 1 to n-1, accumulate the values in the sum[] array as: sum[i] += sum[i-1]. Finally for the index i = 0 to n-1, perform the operation: arr[i] += sum[i].
Implementation:
C++
// C++ implementation to increment values in the // given range by a value d for multiple queries #include <bits/stdc++.h> using namespace std; // structure to store the (start, end) index pair for // each query struct query { int start, end; }; // function to increment values in the given range // by a value d for multiple queries void incrementByD( int arr[], struct query q_arr[], int n, int m, int d) { int sum[n]; memset (sum, 0, sizeof (sum)); // for each (start, end) index pair perform the // following operations on 'sum[]' for ( int i = 0; i < m; i++) { // increment the value at index 'start' by // the given value 'd' in 'sum[]' sum[q_arr[i].start] += d; // if the index '(end+1)' exists then decrement // the value at index '(end+1)' by the given // value 'd' in 'sum[]' if ((q_arr[i].end + 1) < n) sum[q_arr[i].end + 1] -= d; } // Now, perform the following operations: // accumulate values in the 'sum[]' array and // then add them to the corresponding indexes // in 'arr[]' arr[0] += sum[0]; for ( int i = 1; i < n; i++) { sum[i] += sum[i - 1]; arr[i] += sum[i]; } } // function to print the elements of the given array void printArray( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; } // Driver program to test above int main() { int arr[] = { 3, 5, 4, 8, 6, 1 }; struct query q_arr[] = { { 0, 3 }, { 4, 5 }, { 1, 4 }, { 0, 1 }, { 2, 5 } }; int n = sizeof (arr) / sizeof (arr[0]); int m = sizeof (q_arr) / sizeof (q_arr[0]); int d = 2; cout << "Original Array:\n" ; printArray(arr, n); // modifying the array for multiple queries incrementByD(arr, q_arr, n, m, d); cout << "\nModified Array:\n" ; printArray(arr, n); return 0; } |
Java
// Java implementation to increment values in the // given range by a value d for multiple queries class GFG { // structure to store the (start, end) // index pair for each query static class query { int start, end; query( int start, int end) { this .start = start; this .end = end; } } // function to increment values in the given range // by a value d for multiple queries public static void incrementByD( int [] arr, query[] q_arr, int n, int m, int d) { int [] sum = new int [n]; // for each (start, end) index pair, perform // the following operations on 'sum[]' for ( int i = 0 ; i < m; i++) { // increment the value at index 'start' // by the given value 'd' in 'sum[]' sum[q_arr[i].start] += d; // if the index '(end+1)' exists then // decrement the value at index '(end+1)' // by the given value 'd' in 'sum[]' if ((q_arr[i].end + 1 ) < n) sum[q_arr[i].end + 1 ] -= d; } // Now, perform the following operations: // accumulate values in the 'sum[]' array and // then add them to the corresponding indexes // in 'arr[]' arr[ 0 ] += sum[ 0 ]; for ( int i = 1 ; i < n; i++) { sum[i] += sum[i - 1 ]; arr[i] += sum[i]; } } // function to print the elements of the given array public static void printArray( int [] arr, int n) { for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); } // Driver code public static void main(String[] args) { int [] arr = { 3 , 5 , 4 , 8 , 6 , 1 }; query[] q_arr = new query[ 5 ]; q_arr[ 0 ] = new query( 0 , 3 ); q_arr[ 1 ] = new query( 4 , 5 ); q_arr[ 2 ] = new query( 1 , 4 ); q_arr[ 3 ] = new query( 0 , 1 ); q_arr[ 4 ] = new query( 2 , 5 ); int n = arr.length; int m = q_arr.length; int d = 2 ; System.out.println( "Original Array:" ); printArray(arr, n); // modifying the array for multiple queries incrementByD(arr, q_arr, n, m, d); System.out.println( "\nModified Array:" ); printArray(arr, n); } } // This code is contributed by // sanjeev2552 |
Python3
# Python3 implementation to increment # values in the given range by a value d # for multiple queries # structure to store the (start, end) # index pair for each query # function to increment values in the given range # by a value d for multiple queries def incrementByD(arr, q_arr, n, m, d): sum = [ 0 for i in range (n)] # for each (start, end) index pair perform # the following operations on 'sum[]' for i in range (m): # increment the value at index 'start' # by the given value 'd' in 'sum[]' sum [q_arr[i][ 0 ]] + = d # if the index '(end+1)' exists then decrement # the value at index '(end+1)' by the given # value 'd' in 'sum[]' if ((q_arr[i][ 1 ] + 1 ) < n): sum [q_arr[i][ 1 ] + 1 ] - = d # Now, perform the following operations: # accumulate values in the 'sum[]' array and # then add them to the corresponding indexes # in 'arr[]' arr[ 0 ] + = sum [ 0 ] for i in range ( 1 , n): sum [i] + = sum [i - 1 ] arr[i] + = sum [i] # function to print the elements # of the given array def printArray(arr, n): for i in arr: print (i, end = " " ) # Driver Code arr = [ 3 , 5 , 4 , 8 , 6 , 1 ] q_arr = [[ 0 , 3 ], [ 4 , 5 ], [ 1 , 4 ], [ 0 , 1 ], [ 2 , 5 ]] n = len (arr) m = len (q_arr) d = 2 print ( "Original Array:" ) printArray(arr, n) # modifying the array for multiple queries incrementByD(arr, q_arr, n, m, d) print ( "\nModified Array:" ) printArray(arr, n) # This code is contributed # by Mohit Kumar |
C#
// C# implementation to increment values in the // given range by a value d for multiple queries using System; class GFG { // structure to store the (start, end) // index pair for each query public class query { public int start, end; public query( int start, int end) { this .start = start; this .end = end; } } // function to increment values in the given range // by a value d for multiple queries public static void incrementByD( int [] arr, query[] q_arr, int n, int m, int d) { int [] sum = new int [n]; // for each (start, end) index pair, perform // the following operations on 'sum[]' for ( int i = 0; i < m; i++) { // increment the value at index 'start' // by the given value 'd' in 'sum[]' sum[q_arr[i].start] += d; // if the index '(end+1)' exists then // decrement the value at index '(end+1)' // by the given value 'd' in 'sum[]' if ((q_arr[i].end + 1) < n) sum[q_arr[i].end + 1] -= d; } // Now, perform the following operations: // accumulate values in the 'sum[]' array and // then add them to the corresponding indexes // in 'arr[]' arr[0] += sum[0]; for ( int i = 1; i < n; i++) { sum[i] += sum[i - 1]; arr[i] += sum[i]; } } // function to print the elements of the given array public static void printArray( int [] arr, int n) { for ( int i = 0; i < n; i++) Console.Write(arr[i] + " " ); } // Driver code public static void Main(String[] args) { int [] arr = { 3, 5, 4, 8, 6, 1 }; query[] q_arr = new query[5]; q_arr[0] = new query(0, 3); q_arr[1] = new query(4, 5); q_arr[2] = new query(1, 4); q_arr[3] = new query(0, 1); q_arr[4] = new query(2, 5); int n = arr.Length; int m = q_arr.Length; int d = 2; Console.WriteLine( "Original Array:" ); printArray(arr, n); // modifying the array for multiple queries incrementByD(arr, q_arr, n, m, d); Console.WriteLine( "\nModified Array:" ); printArray(arr, n); } } // This code is contributed by Princi Singh |
Javascript
<script> // Javascript implementation to increment values in the // given range by a value d for multiple queries // structure to store the (start, end) // index pair for each query class query { constructor(start,end) { this .start = start; this .end = end; } } // function to increment values in the given range // by a value d for multiple queries function incrementByD(arr,q_arr,n,m,d) { let sum = new Array(n); for (let i=0;i<sum.length;i++) { sum[i]=0; } // for each (start, end) index pair, perform // the following operations on 'sum[]' for (let i = 0; i < m; i++) { // increment the value at index 'start' // by the given value 'd' in 'sum[]' sum[q_arr[i].start] += d; // if the index '(end+1)' exists then // decrement the value at index '(end+1)' // by the given value 'd' in 'sum[]' if ((q_arr[i].end + 1) < n) sum[q_arr[i].end + 1] -= d; } // Now, perform the following operations: // accumulate values in the 'sum[]' array and // then add them to the corresponding indexes // in 'arr[]' arr[0] += sum[0]; for (let i = 1; i < n; i++) { sum[i] += sum[i - 1]; arr[i] += sum[i]; } } // function to print the elements of the given array function printArray(arr,n) { for (let i = 0; i < n; i++) document.write(arr[i] + " " ); } // Driver code let arr = [3, 5, 4, 8, 6, 1 ]; let q_arr = new Array(5); q_arr[0] = new query(0, 3); q_arr[1] = new query(4, 5); q_arr[2] = new query(1, 4); q_arr[3] = new query(0, 1); q_arr[4] = new query(2, 5); let n = arr.length; let m = q_arr.length; let d = 2; document.write( "Original Array:<br>" ); printArray(arr, n); // modifying the array for multiple queries incrementByD(arr, q_arr, n, m, d); document.write( "<br>Modified Array:<br>" ); printArray(arr, n); // This code is contributed by patel2127 </script> |
Original Array: 3 5 4 8 6 1 Modified Array: 7 11 10 14 12 5
Time Complexity: O(m+n)
Auxiliary Space: O(n)