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

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 2

Input: arr[] = { 2, 5, 7, 1, 6, 9, 12, 4, 6 } and x = 30
Output: 5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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 target value lies between ‘position’ and ‘position’ + 2^i, so try to lift by a lower power of 2. Finally loop will end such that 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 using namespace std;    // function to make prefix sums array void MakePreSum(int arr[], int presum[], int n) {     presum = arr;     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) {     // intisalize 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)         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<

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 = arr;     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) {     // intisalize 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)         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<

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 = arr     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):        # intisalize 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):         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<

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 = arr;         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)     {         // intisalize 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)             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<

Output:

2

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

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