Minimum count of indices to be skipped for every index of Array to keep sum till that index at most T

• Last Updated : 16 Jul, 2021

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 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 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 approach class 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 count = new TreeMap();     // 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 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.

Javascript


Output
0 0 0 0 0 2 3

Time Complexity: O(N2)
Auxiliary Space: O(N)

My Personal Notes arrow_drop_up