Network Network Consolidator 19 — Problem Statement & Solution Guide
Problem Description
You are given an array nums of length n. For each index i, locate the first element to the right of i that is strictly greater than nums[i]. If such an element exists, denote it as nextGreater[i]; otherwise, treat nextGreater[i] as 0. Compute the bitwise AND of nums[i] and nextGreater[i] for every position and output the sum of all these AND results.
Input: The first line contains an integer n (1 ≤ n ≤ 2·10^5). The second line contains n space‑separated integers nums[i] (0 ≤ nums[i] ≤ 10^9).
Output: A single integer – the sum of (nums[i] & nextGreater[i]) over all i.
The required algorithm must run in O(n) time and O(n) additional memory, which can be achieved with a monotonic decreasing stack to find the next greater element for each position.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Consolidator 19"
WHY DOES IT MATTER?
Next greater element patterns turn quadratic scans into linear passes, crucial for high‑throughput data processing.
OPTIMIZATION CHALLENGE
The key is reducing repeated comparisons by using a stack to remember only relevant candidates.
REAL-WORLD CONNECTION
Similar logic appears in stock span calculations and real‑time sensor threshold alerts where the next higher reading triggers an action.
Initialize the stack with a sentinel and process the array backwards to avoid extra boundary checks.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem combines the classic Next Greater Element (NGE) pattern with a bitwise aggregation. By scanning the array from right to left and maintaining a monotonic decreasing stack, we can determine for each index the first greater value to its right in O(n) time, which is essential because a naive double loop would require O(n^2) comparisons and time out on large inputs. Once the NGE array is built, the final step is a linear pass that computes (nums[i] & nextGreater[i]) and accumulates the sum, leveraging the constant‑time nature of bitwise AND. This two‑phase approach—first resolve the relational structure with a stack, then perform a simple arithmetic reduction—embodies the optimal paradigm for problems that ask for a nearest‑greater relationship followed by a per‑element calculation.
Interview Questions on This Problem
Q1How does a monotonic stack help find the next greater element in linear time?
The stack stores candidates in decreasing order, so when a larger element appears it resolves all smaller elements to its left at once. Each array element is pushed and popped at most once, guaranteeing O(n) operations.
Q2Why is the bitwise AND operation safe to compute after finding next greater values?
AND is a constant‑time, associative operation that does not depend on other indices, so it can be applied independently per position. This decouples the aggregation from the NGE discovery, preserving linear complexity.
Q3What edge cases must be handled when the next greater element does not exist?
If no greater element exists, nextGreater[i] is defined as 0, and (nums[i] & 0) yields 0, contributing nothing to the sum. Ensure the algorithm explicitly sets missing NGEs to 0 rather than leaving them undefined.
Examples
Input
5 5 1 3 4 2
Output
1
Explanation: Next greater elements: [0,3,4,0,0]. ANDs: 5&0=0, 1&3=1, 3&4=0, 4&0=0, 2&0=0. Sum = 1.
Input
6 2 7 3 9 5 8
Output
4
Explanation: Next greater elements: [7,9,9,0,8,0]. ANDs: 2&7=2, 7&9=1, 3&9=1, 9&0=0, 5&8=0, 8&0=0. Sum = 2+1+1 = 4.
Input
4 10 6 12 4
Output
12
Explanation: Next greater elements: [12,12,0,0]. ANDs: 10&12=8, 6&12=4, 12&0=0, 4&0=0. Sum = 8+4 = 12.
Constraints
- 1 <= n <= 2*10^5
- 0 <= nums[i] <= 10^9
- All calculations fit in 64‑bit signed integer
- Algorithm must run in O(n) time
Optimal Approach & Strategy
Traverse the array from right to left with a monotonic decreasing stack to obtain next greater values in O(n), then sum the ANDs in a second linear pass.
Brute Force Approach
Iterate over each index i and scan all positions j > i until a larger value is found, then compute the AND; this is O(n^2) and too slow for large n.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (typeof num === 'number' && num > K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (typeof num === 'number' && num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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