Skip to content
Related Articles

Related Articles

Improve Article

First element greater than or equal to X in prefix sum of N numbers using Binary Lifting

  • Last Updated : 23 Jun, 2021

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:
prefix sum array formed is { 2, 7, 14, 15, 21, 30, 42, 46, 52}, hence 14 is the number whose index is 2

Input: 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:  

C++




// 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




// 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




# 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#




// 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

Javascript




<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>
Output: 
2

 

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :