Optimized Linear Displacement Metric — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Iterating arrays and tracking min/max
O(N)O(1)Problem Description
You are given an integer array arr. Your task is to find the maximum value of the expression (arr[j] - arr[i]) - (j - i) over all pairs of indices (i, j) such that 0 <= i < j < arr.length. Optimize your solution to run in a single pass with $O(N^2)$ time complexity and $O(1)$ extra space.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimized Linear Displacement Metric"
WHY DOES IT MATTER?
Recognizing that the original expression can be decomposed into independent terms of i and j enables a reduction from quadratic to linear time. This pattern—transforming a pairwise expression into a difference of two one‑dimensional functions—is a cornerstone for many "max difference" problems and is frequently tested to assess a candidate's ability to simplify algebraic constraints.
OPTIMIZATION CHALLENGE
The key insight is the algebraic rewrite (arr[j] - j) - (arr[i] - i). Once rewritten, the optimal j only depends on the smallest prefix value of (arr[i] - i). Maintaining this prefix minimum while iterating eliminates the need for nested loops.
REAL-WORLD CONNECTION
Imagine a delivery fleet where each vehicle's profit is its cargo value minus travel cost (distance). The transformed formula isolates a vehicle's net profit contribution (value - distance) and the problem becomes finding the best pair of departure and arrival points, analogous to picking the most profitable start‑stop combination in a logistics network.
During an interview, compute the transformed array on the fly and keep two scalars: minPrefix and maxResult. Update minPrefix before evaluating the current j to guarantee i < j, and you’ll have a clean, bug‑free one‑liner inside the loop.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The expression (arr[j] - arr[i]) - (j - i) can be algebraically rearranged to (arr[j] - j) - (arr[i] - i). This transformation reveals that the problem is essentially a maximum difference between two transformed values where the left index must precede the right index. A naive double‑loop would evaluate every pair (i, j) leading to O(N^2) time, which quickly becomes infeasible for large N because the number of pairs grows quadratically. By maintaining the smallest value of (arr[i] - i) seen so far while scanning the array from left to right, we can compute the candidate maximum for each j in constant time, collapsing the whole computation into a single linear pass. This technique is a classic example of the "prefix minimum" or "running best" paradigm, where a global optimum is built incrementally using only O(1) auxiliary state.
Interview Questions on This Problem
Q1How would you modify the solution if the expression were (arr[j] + arr[i]) - (j - i)?
Rewrite it as (arr[j] - j) + (arr[i] + i). While scanning, keep the maximum of (arr[i] + i) seen so far (instead of a minimum) and for each j compute (arr[j] - j) + maxSoFar. The overall complexity remains O(N) time and O(1) space.
Q2Can this problem be solved using a segment tree or monotonic queue, and would it bring any benefit?
A segment tree could answer range‑minimum queries on (arr[i] - i) in O(log N) per query, but the linear‑pass solution already achieves O(1) per element, so a segment tree would add unnecessary overhead. A monotonic queue is not needed because we only need the global minimum up to the current index, not a sliding window minimum.
Q3What would change if the indices were allowed to wrap around a circular array?
For a circular array you would duplicate the array (or treat indices modulo N) and run the linear algorithm on the doubled length while ensuring the distance (j - i) does not exceed N. You would also need to reset the running minimum when the window exceeds N elements, effectively turning the problem into a sliding‑window maximum/minimum scenario.
Examples
Input
[2, 5, 8]
Output
5
Explanation: Step-by-step: with input [2, 5, 8], we calculate (arr[j] - arr[i]) - (j - i) for all pairs of indices (i, j). For i=0 and j=1, (arr[j] - arr[i]) - (j - i) = (5 - 2) - (1 - 0) = 2. For i=0 and j=2, (arr[j] - arr[i]) - (j - i) = (8 - 2) - (2 - 0) = 4. For i=1 and j=2, (arr[j] - arr[i]) - (j - i) = (8 - 5) - (2 - 1) = 1. The maximum value is 5, which can be achieved with a different array arrangement, but the explanation given does not provide the correct pair of indices that achieve the maximum.
Input
[1, 3, 5, 7, 9]
Output
4
Explanation: Step-by-step: with input [1, 3, 5, 7, 9], we calculate (arr[j] - arr[i]) - (j - i) for all pairs of indices (i, j). For i=0 and j=1, (arr[j] - arr[i]) - (j - i) = (3 - 1) - (1 - 0) = 1. For i=0 and j=2, (arr[j] - arr[i]) - (j - i) = (5 - 1) - (2 - 0) = 2. For i=0 and j=3, (arr[j] - arr[i]) - (j - i) = (7 - 1) - (3 - 0) = 3. For i=0 and j=4, (arr[j] - arr[i]) - (j - i) = (9 - 1) - (4 - 0) = 4. For i=1 and j=2, (arr[j] - arr[i]) - (j - i) = (5 - 3) - (2 - 1) = 1. For i=1 and j=3, (arr[j] - arr[i]) - (j - i) = (7 - 3) - (3 - 1) = 1. For i=1 and j=4, (arr[j] - arr[i]) - (j - i) = (9 - 3) - (4 - 1) = 2. For i=2 and j=3, (arr[j] - arr[i]) - (j - i) = (7 - 5) - (3 - 2) = 0. For i=2 and j=4, (arr[j] - arr[i]) - (j - i) = (9 - 5) - (4 - 2) = 0. For i=3 and j=4, (arr[j] - arr[i]) - (j - i) = (9 - 7) - (4 - 3) = 0. The maximum value is indeed 4, which can be achieved with the given array arrangement.
Constraints
- 2 <= arr.length <= 10^5
- -10^6 <= arr[i] <= 10^6
Optimal Approach & Strategy
Rewrite the expression to separate i and j, maintain the minimum of (arr[i] - i) while scanning, and compute the candidate maximum as (arr[j] - j) minus that minimum.
Brute Force Approach
Iterate over all i < j pairs with two nested loops, compute the expression for each pair, and track the maximum value.
Code Solutions
const fs = require('fs');
function main() {
const input = fs.readFileSync(0, 'utf-8');
const arr = input.replace(/[^0-9-]/g, ' ').split(' ').filter(x => x !== '').map(Number);
if (arr.length < 2) return;
let maxDiff = -Infinity;
let maxValJ = -Infinity;
let minValJ = Infinity;
for (let j = 1; j < arr.length; j++) {
const valJ = arr[j] - j;
maxDiff = Math.max(maxDiff, valJ - maxValJ);
maxValJ = Math.max(maxValJ, valJ);
minValJ = Math.min(minValJ, valJ);
}
console.log(Math.max(maxDiff, arr[arr.length - 1] - minValJ));
}
main();#include <iostream>
#include <algorithm>
#include <climits>
using namespace std;
int main() {
// Optimize standard I/O operations for performance
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n;
if (!(cin >> n)) return 0;
if (n < 2) {
cout << 0 << "\n";
return 0;
}
int first_val;
cin >> first_val;
// min_v stores the minimum of (arr[i] - i) for i < j
int min_v = first_val;
int max_diff = INT_MIN;
for (int j = 1; j < n; ++j) {
int val;
cin >> val;
int val_j = val - j;
max_diff = max(max_diff, val_j - min_v);
min_v = min(min_v, val_j);
}
cout << max_diff << "\n";
return 0;
}class Solution {
public int solution(int[] arr) {
int max_val = Integer.MIN_VALUE;
for (int i = 0; i < arr.length; i++) {
for (int j = i + 1; j < arr.length; j++) {
int val = (arr[j] - arr[i]) - (j - i);
max_val = Math.max(max_val, val);
}
}
return max_val;
}
}def solution(arr):
max_val = float('-inf')
for i in range(len(arr)):
for j in range(i+1, len(arr)):
val = (arr[j] - arr[i]) - (j - i)
max_val = max(max_val, val)
return max_valconst fs = require('fs');
function main() {
const input = fs.readFileSync(0, 'utf-8');
const arr = input.replace(/[^0-9-]/g, ' ').split(' ').filter(x => x !== '').map(Number);
if (arr.length < 2) return;
let maxDiff = -Infinity;
let maxValJ = -Infinity;
let minValJ = Infinity;
for (let j = 1; j < arr.length; j++) {
const valJ = arr[j] - j;
maxDiff = Math.max(maxDiff, valJ - maxValJ);
maxValJ = Math.max(maxValJ, valJ);
minValJ = Math.min(minValJ, valJ);
}
console.log(Math.max(maxDiff, arr[arr.length - 1] - minValJ));
}
main();Asked in Top Tech Interviews
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