Given an array, arr[] of size N and an integer T. The task is to find for each index the minimum number of indices that should be skipped if the sum till the ith index should not exceed T.
Examples:
Input: N = 7, T = 15, arr[] = {1, 2, 3, 4, 5, 6, 7}
Output: 0 0 0 0 0 2 3
Explanation: No indices need to be skipped for the first 5 indices: {1, 2, 3, 4, 5}, since their sum is 15 and <= T.
For the sixth index, indices 3 and 4 can be skipped, that makes its sum = (1+2+3+6) = 12.
For the seventh, indices 3, 4 and 5 can be skipped which makes its sum = (1+2+3+7) = 13.Input: N = 2, T = 100, arr[] = {100, 100}
Output: 0 1
Approach: The idea is to use a map to store the visited elements in increasing order while traversing. Follow the steps below to solve the problem:
- Create an ordered map, M to keep a count of the elements before the ith index.
- Initialize a variable sum as 0 to store the prefix sum.
- Traverse the array, arr[] using the variable i
- Store the difference of sum+arr[i] and T in a variable, d.
- If the value of d>0, traverse the map from the end and select the indices with the largest elements until the sum becomes less than T. Store the number of elements required in a variable k.
- Add arr[i] to sum and increment A[i] in M by 1.
- Print the value of k.
Below is the implementation of the above approach:
C++
// C++ approach for above approach#include <bits/stdc++.h>using namespace std;// Function to calculate minimum indices to be skipped// so that sum till i remains smaller than Tvoid skipIndices(int N, int T, int arr[]){ // Store the sum of all indices before i int sum = 0; // Store the elements that can be skipped map<int, int> count; // Traverse the array, A[] for (int i = 0; i < N; i++) { // Store the total sum of elements that // needs to be skipped int d = sum + arr[i] - T; // Store the number of elements need // to be removed int k = 0; if (d > 0) { // Traverse from the back of map so // as to take bigger elements first for (auto u = count.rbegin(); u != count.rend(); u++) { int j = u->first; int x = j * count[j]; if (d <= x) { k += (d + j - 1) / j; break; } k += count[j]; d -= x; } } // Update sum sum += arr[i]; // Update map with the current element count[arr[i]]++; cout << k << " "; }}// Driver codeint main(){ // Given Input int N = 7; int T = 15; int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; // Function Call skipIndices(N, T, arr); return 0;} |
Java
import java.util.ArrayList;import java.util.Map;import java.util.TreeMap;// C++ approach for above approachclass GFG { // Function to calculate minimum indices to be skipped // so that sum till i remains smaller than Tpublic static void skipIndices(int N, int T, int arr[]){ // Store the sum of all indices before i int sum = 0; // Store the elements that can be skipped TreeMap<Integer, Integer> count = new TreeMap<Integer, Integer>(); // Traverse the array, A[] for (int i = 0; i < N; i++) { // Store the total sum of elements that // needs to be skipped int d = sum + arr[i] - T; // Store the number of elements need // to be removed int k = 0; if (d > 0) { // Traverse from the back of map so // as to take bigger elements first for (Map.Entry<Integer, Integer> u : count.descendingMap().entrySet()) { int j = u.getKey(); int x = j * count.get(j); if (d <= x) { k += (d + j - 1) / j; break; } k += count.get(j); d -= x; } } // Update sum sum += arr[i]; // Update map with the current element if(count.containsKey(arr[i])){ count.put(arr[i], count.get(arr[i]) + 1); }else{ count.put(arr[i], 1); } System.out.print(k + " "); }} // Driver code public static void main(String args[]) { // Given Input int N = 7; int T = 15; int arr[] = { 1, 2, 3, 4, 5, 6, 7 }; // Function Call skipIndices(N, T, arr); }}// This code is contributed by _saurabh_jaiswal. |
Python3
# Python3 approach for above approach# Function to calculate minimum indices to be skipped# so that sum till i remains smaller than Tdef skipIndices(N, T, arr): # Store the sum of all indices before i sum = 0 # Store the elements that can be skipped count = {} # Traverse the array, A[] for i in range(N): # Store the total sum of elements that # needs to be skipped d = sum + arr[i] - T # Store the number of elements need # to be removed k = 0 if (d > 0): # Traverse from the back of map so # as to take bigger elements first for u in list(count.keys())[::-1]: j = u x = j * count[j] if (d <= x): k += (d + j - 1) // j break k += count[j] d -= x # Update sum sum += arr[i] # Update map with the current element count[arr[i]] = count.get(arr[i], 0) + 1 print(k, end = " ")# Driver codeif __name__ == '__main__': # Given Input N = 7 T = 15 arr = [1, 2, 3, 4, 5, 6, 7] # Function Call skipIndices(N, T, arr) # This code is contributed by mohit kumar 29. |
C#
using System;using System.Collections.Generic;using System.Linq;class GFG{ // Function to calculate minimum indices to be skipped // so that sum till i remains smaller than T public static void skipIndices(int N, int T, int[] arr) { // Store the sum of all indices before i int sum = 0; // Store the elements that can be skipped SortedDictionary<int, int> count = new SortedDictionary<int, int>(); // Traverse the array, A[] for (int i = 0; i < N; i++) { // Store the total sum of elements that // needs to be skipped int d = sum + arr[i] - T; // Store the number of elements need // to be removed int k = 0; if (d > 0) { // Traverse from the back of map so // as to take bigger elements first foreach (KeyValuePair<int, int> u in count.Reverse()) { int j = u.Key; int x = j * count[j]; if (d <= x) { k += (d + j - 1) / j; break; } k += count[j]; d -= x; } } // Update sum sum += arr[i]; // Update map with the current element if (count.ContainsKey(arr[i])) { count[arr[i]] = count[arr[i]] + 1; } else { count[arr[i]] = 1; } Console.Write(k + " "); } } // Driver code public static void Main(string[] args) { // Given Input int N = 7; int T = 15; int[] arr = { 1, 2, 3, 4, 5, 6, 7 }; // Function Call skipIndices(N, T, arr); }} |
Javascript
<script>// Javascript approach for above approach// Function to calculate minimum indices // to be skipped so that sum till i // remains smaller than Tfunction skipIndices(N, T, arr) { // Store the sum of all indices before i let sum = 0; // Store the elements that can be skipped let count = new Map(); // Traverse the array, A[] for(let i = 0; i < N; i++) { // Store the total sum of elements that // needs to be skipped let d = sum + arr[i] - T; // Store the number of elements need // to be removed let k = 0; if (d > 0) { // Traverse from the back of map so // as to take bigger elements first for(let u of [...count.keys()].reverse()) { let j = u; let x = j * count.get(j); if (d <= x) { k += Math.floor((d + j - 1) / j); break; } k += count.get(j); d -= x; } } // Update sum sum += arr[i]; // Update map with the current element if (count.has(arr[i])) { count.set(arr[i], count.get(i) + 1) } else { count.set(arr[i], 1) } document.write(k + " "); }}// Driver code// Given Inputlet N = 7;let T = 15;let arr = [ 1, 2, 3, 4, 5, 6, 7 ];// Function CallskipIndices(N, T, arr);// This code is contributed by _saurabh_jaiswal</script> |
Output
0 0 0 0 0 2 3
Time Complexity: O(N2)
Auxiliary Space: O(N)
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