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++ 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 T 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
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 code int 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;
} |
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 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
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 approach for above approach # Function to calculate minimum indices to be skipped # so that sum till i remains smaller than T def 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 code if __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.
|
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);
}
} |
<script> // Javascript approach for above approach // Function to calculate minimum indices // to be skipped so that sum till i // remains smaller than T function 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 Input let N = 7; let T = 15; let arr = [ 1, 2, 3, 4, 5, 6, 7 ]; // Function Call skipIndices(N, T, arr); // This code is contributed by _saurabh_jaiswal </script> |
0 0 0 0 0 2 3
Time Complexity: O(N2)
Auxiliary Space: O(N)