Balanced Pointer Alignment — Problem Statement & Solution Guide

StackMediumNext Greater Element

TimeO(n)

SpaceO(n)

Problem Description

Given an array of integers, compute the sum of differences between each element and its next greater element according to the Next Greater Element algorithm. The next greater element for an element is the smallest element on its right that is greater than the element.

DSA Pattern Breakdown

"Balanced Pointer Alignment"

medium

WHY DOES IT MATTER?

Monotonic stacks provide an elegant, linear-time solution to a class of problems where each element’s answer depends on a future element that satisfies a monotonic condition. They avoid nested loops and reduce time complexity from quadratic to linear, which is critical for large datasets.

OPTIMIZATION CHALLENGE

The key insight is that once an element’s NGE is found, it never needs to be considered again. By popping elements from the stack when a greater element is found, we eliminate future comparisons and achieve O(n) time.

REAL-WORLD CONNECTION

Consider a real-time monitoring system that tracks server load spikes. Each spike’s impact is only relevant until a higher spike occurs; a monotonic stack efficiently tracks the next higher spike, analogous to the NGE algorithm.

When explaining this pattern, emphasize the stack’s invariant (decreasing values) and how it guarantees that each element is processed exactly twice—once when pushed and once when popped.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(n)

Core Theory — Why This Approach?

The Next Greater Element (NGE) problem asks for each array element to find the first element to its right that is larger. A naive solution scans rightwards for every element, leading to O(n^2) time and is infeasible for large inputs (e.g., n=10^6). The optimal solution uses a monotonic stack: we traverse the array from left to right, maintaining a stack of indices whose NGE has not yet been found. When we encounter a value greater than the stack’s top, we pop and record that value as the NGE for the popped index. This guarantees each element is pushed and popped at most once, yielding linear time.

The stack’s monotonic property (strictly decreasing values from bottom to top) ensures that once an element is popped, all smaller elements to its left cannot have a greater element beyond the current index, so their NGE is already determined. This eliminates redundant comparisons and reduces the problem to a single pass. The space complexity is O(n) for the stack and the result array, but the stack size never exceeds n.

In practice, this pattern is a cornerstone for many “next/previous” problems (e.g., stock span, histogram area). Understanding the stack’s role in maintaining a candidate set of indices is key to mastering efficient solutions for large-scale data streams and real-time analytics.

Interview Questions on This Problem

Q1How would you modify the algorithm if the array contains duplicate values and you need the next strictly greater element?

The algorithm remains unchanged; the stack condition uses a strict '>' comparison. Duplicates are treated as not greater, so they stay on the stack until a strictly larger value appears.

Q2In a distributed system, how could you parallelize the NGE computation for a large array split across nodes?

Partition the array into chunks, compute local NGEs, then propagate the last element of each chunk to the next node to resolve cross-boundary greater elements. A merge step reconciles boundary NGEs using a stack that spans the partition borders.

Q3What is the time complexity if you replace the stack with a balanced binary search tree for the NGE problem?

Using a BST would increase each operation to O(log n), leading to O(n log n) overall, which is worse than the O(n) stack approach. The stack’s linear behavior is due to its amortized constant-time push/pop operations.

Examples

Example 1

Input

[8, 6, 7, 9, 3, 1]

Output

Explanation: Step-by-step: 1. Initialize sum to 0. 2. Iterate through the array from left to right. 3. For each element, find its next greater element. 4. If the next greater element exists, add the difference between the current element and the next greater element to the sum. 5. Return the sum.

Example 2

Input

[1, 2, 3, 4, 5]

Output

Explanation: Step-by-step: 1. Initialize sum to 0. 2. Iterate through the array from left to right. 3. For each element, find its next greater element. 4. Since there is no next greater element for any element, return the sum as 0.

Constraints

1 <= N <= 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity expected: O(N) or O(N log N)
Space Complexity expected: O(1) or O(N)

Optimal Approach & Strategy

Traverse the array once while maintaining a decreasing stack of indices. When a larger element is found, pop indices and record the difference between the current element and the popped element’s value. This achieves O(n) time and O(n) space.

Brute Force Approach

For each element, scan all elements to its right until you find a greater one, then compute the difference. This takes O(n^2) time and is impractical for large arrays.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function solution(nums) {
   if (nums.length <= 1) return 0;
   let stack = [];
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       while (stack.length > 0 && nums[stack[stack.length - 1]] < nums[i]) {
           let j = stack.pop();
           sum += nums[i] - nums[j];
       }
       stack.push(i);
   }
   return sum;
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       if (nums.size() <= 1) return 0;
       vector<int> stack;
       int sum = 0;
       for (int i = 0; i < nums.size(); i++) {
           while (!stack.empty() && nums[stack.back()] < nums[i]) {
               int j = stack.back();
               stack.pop_back();
               sum += nums[i] - nums[j];
           }
           stack.push_back(i);
       }
       return sum;
   }
}

class Solution {
   public int solution(int[] nums) {
       if (nums.length <= 1) return 0;
       int[] stack = new int[nums.length];
       int top = -1;
       int sum = 0;
       for (int i = 0; i < nums.length; i++) {
           while (top >= 0 && nums[stack[top]] < nums[i]) {
               int j = stack[top--];
               sum += nums[i] - nums[j];
           }
           stack[++top] = i;
       }
       return sum;
   }
}

def solution(nums):
   if len(nums) <= 1: return 0
   stack = []
   sum = 0
   for i in range(len(nums)):
       while stack and nums[stack[-1]] < nums[i]:
           j = stack.pop()
           sum += nums[i] - nums[j]
       stack.append(i)
   return sum

function solution(nums) {
   if (nums.length <= 1) return 0;
   let stack = [];
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       while (stack.length > 0 && nums[stack[stack.length - 1]] < nums[i]) {
           let j = stack.pop();
           sum += nums[i] - nums[j];
       }
       stack.push(i);
   }
   return sum;
}

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Balanced Pointer Alignment Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Stack

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Balanced Pointer Alignment — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor