Given an array of N integers and a number X. The task is to find the index of first element which is greater than or equal to X in prefix sums of N numbers.
Examples:
Input: arr[] = { 2, 5, 7, 1, 6, 9, 12, 4, 6 } and x = 8
Output: 2
prefix sum array formed is { 2, 7, 14, 15, 21, 30, 42, 46, 52}, hence 14 is the number whose index is 2Input: arr[] = { 2, 5, 7, 1, 6, 9, 12, 4, 6 } and x = 30
Output: 5
Approach: The problem can be solved using lower_bound function in Binary search. But in this post, the problem will be solved using Binary-Lifting. In binary lifting, a value is increased (or lifted) by powers of 2, starting with the highest possible power of 2(log(N)) down to the lowest power(0).
- Initialize position = 0 and set each bit of position, from most significant bit to least significant bit.
- Whenever a bit is set to 1, the value of position increases (or lifts).
- While increasing or lifting position, make sure that prefix sum till position should be less than v.
- Here, log(N) bits are needed for all possible values of ‘position’ ( from log(N)th to 0th bit ).
- Determine the value of the i-th bit. First, check if setting the i-th bit won’t make ‘position’ greater than N, which is the size of the array. Before lifting to the new ‘position’, check that value at that new ‘position’ is less than X or not.
- If this condition is true, then target position lies above the ‘position’ + 2^i, but below the ‘position’ + 2^(i+1). This is because if the target position was above ‘position’ + 2^(i+1), then the position would have been already lifted by 2^(i+1) (this logic is similar to binary lifting in trees).
- If it is false, then the target value lies between ‘position’ and ‘position’ + 2^i, so try to lift by a lower power of 2. Finally, the loop will end such that the value at that position is less than X. Here, in this question, the lower bound is asked. So, return ‘position’ + 1.
Below is the implementation of the above approach:
// CPP program to find lower_bound of x in // prefix sums array using binary lifting. #include <bits/stdc++.h> using namespace std;
// function to make prefix sums array void MakePreSum( int arr[], int presum[], int n)
{ presum[0] = arr[0];
for ( int i = 1; i < n; i++)
presum[i] = presum[i - 1] + arr[i];
} // function to find lower_bound of x in // prefix sums array using binary lifting. int BinaryLifting( int presum[], int n, int x)
{ // initialize position
int pos = 0;
// find log to the base 2 value of n.
int LOGN = log2(n);
// if x less than first number.
if (x <= presum[0])
return 0;
// starting from most significant bit.
for ( int i = LOGN; i >= 0; i--) {
// if value at this position less
// than x then updateposition
// Here (1<<i) is similar to 2^i.
if (pos + (1 << i) < n &&
presum[pos + (1 << i)] < x) {
pos += (1 << i);
}
}
// +1 because 'pos' will have position
// of largest value less than 'x'
return pos + 1;
} // Driver code int main()
{ // given array
int arr[] = { 2, 5, 7, 1, 6, 9, 12, 4, 6 };
// value to find
int x = 8;
// size of array
int n = sizeof (arr) / sizeof (arr[0]);
// to store prefix sum
int presum[n] = { 0 };
// call for prefix sum
MakePreSum(arr, presum, n);
// function call
cout << BinaryLifting(presum, n, x);
return 0;
} |
// Java program to find lower_bound of x in // prefix sums array using binary lifting. import java.util.*;
class solution
{ // function to make prefix sums array static void MakePreSum( int []arr, int []presum, int n)
{ presum[ 0 ] = arr[ 0 ];
for ( int i = 1 ; i < n; i++)
presum[i] = presum[i - 1 ] + arr[i];
} // function to find lower_bound of x in // prefix sums array using binary lifting. static int BinaryLifting( int []presum, int n, int x)
{ // initialize position
int pos = 0 ;
// find log to the base 2 value of n.
int LOGN = ( int )Math.log(n);
// if x less than first number.
if (x <= presum[ 0 ])
return 0 ;
// starting from most significant bit.
for ( int i = LOGN; i >= 0 ; i--) {
// if value at this position less
// than x then updateposition
// Here (1<<i) is similar to 2^i.
if (pos + ( 1 << i) < n &&
presum[pos + ( 1 << i)] < x) {
pos += ( 1 << i);
}
}
// +1 because 'pos' will have position
// of largest value less than 'x'
return pos + 1 ;
} // Driver code public static void main(String args[])
{ // given array
int []arr = { 2 , 5 , 7 , 1 , 6 , 9 , 12 , 4 , 6 };
// value to find
int x = 8 ;
// size of array
int n = arr.length;
// to store prefix sum
int []presum = new int [n];
Arrays.fill(presum, 0 );
// call for prefix sum
MakePreSum(arr, presum, n);
// function call
System.out.println(BinaryLifting(presum, n, x));
} } |
# Python 3 program to find # lower_bound of x in prefix # sums array using binary lifting. import math
# function to make prefix # sums array def MakePreSum( arr, presum, n):
presum[ 0 ] = arr[ 0 ]
for i in range ( 1 , n):
presum[i] = presum[i - 1 ] + arr[i]
# function to find lower_bound of x in # prefix sums array using binary lifting. def BinaryLifting(presum, n, x):
# initialize position
pos = 0
# find log to the base 2
# value of n.
LOGN = int (math.log2(n))
# if x less than first number.
if (x < = presum[ 0 ]):
return 0
# starting from most significant bit.
for i in range (LOGN, - 1 , - 1 ) :
# if value at this position less
# than x then updateposition
# Here (1<<i) is similar to 2^i.
if (pos + ( 1 << i) < n and
presum[pos + ( 1 << i)] < x) :
pos + = ( 1 << i)
# +1 because 'pos' will have position
# of largest value less than 'x'
return pos + 1
# Driver code if __name__ = = "__main__" :
# given array
arr = [ 2 , 5 , 7 , 1 , 6 ,
9 , 12 , 4 , 6 ]
# value to find
x = 8
# size of array
n = len (arr)
# to store prefix sum
presum = [ 0 ] * n
# call for prefix sum
MakePreSum(arr, presum, n)
# function call
print (BinaryLifting(presum, n, x))
# This code is contributed # by ChitraNayal |
// C# program to find lower_bound of x in // prefix sums array using binary lifting. using System;
class GFG
{ // function to make prefix sums array
static void MakePreSum( int []arr,
int []presum, int n)
{
presum[0] = arr[0];
for ( int i = 1; i < n; i++)
presum[i] = presum[i - 1] + arr[i];
}
// function to find lower_bound of x in
// prefix sums array using binary lifting.
static int BinaryLifting( int []presum,
int n, int x)
{
// initialize position
int pos = 0;
// find log to the base 2 value of n.
int LOGN = ( int )Math.Log(n);
// if x less than first number.
if (x <= presum[0])
return 0;
// starting from most significant bit.
for ( int i = LOGN; i >= 0; i--)
{
// if value at this position less
// than x then updateposition
// Here (1<<i) is similar to 2^i.
if (pos + (1 << i) < n &&
presum[pos + (1 << i)] < x)
{
pos += (1 << i);
}
}
// +1 because 'pos' will have position
// of largest value less than 'x'
return pos + 1;
}
// Driver code
public static void Main()
{
// given array
int []arr = { 2, 5, 7, 1, 6, 9, 12, 4, 6 };
// value to find
int x = 8;
// size of array
int n = arr.Length;
// to store prefix sum
int []presum = new int [n];
// call for prefix sum
MakePreSum(arr, presum, n);
// function call
Console.WriteLine(BinaryLifting(presum, n, x));
}
} // This code has been contributed // by PrinciRaj1992 |
<script> // Javascript program to find lower_bound of x in // prefix sums array using binary lifting. // function to make prefix sums array
function MakePreSum(arr,presum,n)
{
presum[0] = arr[0];
for (let i = 1; i < n; i++)
presum[i] = presum[i - 1] + arr[i];
}
// function to find lower_bound of x in // prefix sums array using binary lifting. function BinaryLifting(presum, n,k)
{ // initialize position
let pos = 0;
// find log to the base 2 value of n.
let LOGN = Math.log(n);
// if x less than first number.
if (x <= presum[0])
return 0;
// starting from most significant bit.
for (let i = LOGN; i >= 0; i--) {
// if value at this position less
// than x then updateposition
// Here (1<<i) is similar to 2^i.
if (pos + (1 << i) < n &&
presum[pos + (1 << i)] < x) {
pos += (1 << i);
}
}
// +1 because 'pos' will have position
// of largest value less than 'x'
return pos + 1;
} // Driver code // given array let arr=[2, 5, 7, 1, 6, 9, 12, 4, 6]; // value to find let x = 8; // size of array let n = arr.length; // to store prefix sum let presum = new Array(n);
for (let i=0;i<n;i++)
{ presum[i]=0;
} // call for prefix sum MakePreSum(arr, presum, n); // function call document.write(BinaryLifting(presum, n, x)); // This code is contributed by avanitrachhadiya2155 </script> |
2
Time Complexity: O(N)
Auxiliary Space: O(N)