Optimal Stage Index Shifts — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Optimal Stage Index Shifts problem optimally.
O(n log n)O(n)Problem Description
You are given an array arr of n distinct non‑negative integers. The value of each element denotes the original index of a group that should appear on a stage. The desired final arrangement is the array sorted in strictly increasing order. In one operation you may select any element at position i (0 ≤ i < n, i > 0) and insert it at an earlier position j (0 ≤ j < i). The insertion shifts the elements in [j, i‑1] one step to the right. An operation is allowed only if the array remains strictly increasing after the insertion. Determine the minimum number of such operations required to transform arr into a strictly increasing sequence.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Optimal Stage Index Shifts"
WHY DOES IT MATTER?
The LIS pattern captures the maximal set of elements that are already in order, turning a seemingly combinatorial re‑ordering problem into a simple subtraction, which is a recurring theme in many interview problems involving minimum moves.
OPTIMIZATION CHALLENGE
Recognizing that each left‑insert can place an element anywhere earlier eliminates the need to simulate each shift; the challenge is to identify the longest increasing subsequence, which can be found in O(n log n) using binary search on a dynamic tail array.
REAL-WORLD CONNECTION
Think of a production line where items must exit in a specific order; workers can only pull items forward, not push them back. Keeping the longest already‑correct subsequence on the belt minimizes the number of manual pulls, analogous to minimizing left‑shifts in the array.
During the interview, first state the reduction to LIS, then immediately mention the O(n log n) patience‑sorting algorithm—this shows you understand both the problem transformation and the optimal data‑structure technique.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
The problem reduces to finding the minimum number of left‑insert operations required to transform the given permutation into a sorted sequence. Because each operation can only move an element to an earlier position, any element that already appears in the correct relative order with respect to the others can be left untouched; all other elements must be pulled leftwards. This observation maps directly to the classic Longest Increasing Subsequence (LIS) problem: the longest subsequence that is already strictly increasing can stay in place, and every element outside this subsequence must be moved. A naive approach that simulates each possible insertion quickly explodes to O(n^2) or worse, as each move may require shifting many elements. The optimal paradigm leverages the patience‑sorting technique to compute the LIS length in O(n log n) time, allowing us to answer the minimal operation count as n‑LIS. This transformation from a sorting‑by‑shifts problem to LIS is the key theoretical insight that unlocks the efficient solution.
Interview Questions on This Problem
Q1How would you compute the minimum number of left‑insert operations to sort a distinct integer array, and why does the Longest Increasing Subsequence give the answer?
Compute the length of the LIS using the O(n log n) patience‑sorting method; the elements of the LIS are already in correct relative order and need no moves, so the answer is n minus that length.
Q2If the array may contain duplicate values, does the same LIS‑based solution work? Explain any modifications needed.
With duplicates the final sorted order is non‑decreasing, so we need the Longest Non‑Decreasing Subsequence instead; the binary‑search condition changes from < to ≤ when building the piles.
Q3Can you adapt the solution to also output the exact sequence of insert operations required? What additional data structures would you use?
Yes; while computing the LIS keep predecessor indices and a stack of chosen elements, then traverse the array from right to left moving any element not in the LIS to its correct earlier position, recording (i,j) pairs. A Fenwick tree can help locate the target j efficiently.
Examples
Input
[4, 2, 5, 1, 3]
Output
3
Explanation: The sorted order is [1,2,3,4,5]. The longest increasing subsequence that can stay untouched is [2,5] (length 2). All other three elements (4,1,3) must be moved leftwards. One possible sequence of moves: 1) move element 1 (at index 3) to position 0 → [1,4,2,5,3]; 2) move element 3 (at index 4) to position 2 → [1,4,3,2,5]; 3) move element 4 (at index 1) to position 3 → [1,3,2,4,5] which is now sorted after a final left‑shift of 2 and 3. Hence 3 operations are sufficient and optimal.
Input
[0,1,2,3]
Output
0
Explanation: The array is already strictly increasing, so no shift is needed.
Input
[10, 5, 8, 3, 7, 12]
Output
4
Explanation: Sorted target: [3,5,7,8,10,12]. The longest increasing subsequence that respects the original order is [5,8,12] (length 3). Therefore 6‑3 = 3 elements must be moved. However element 10 cannot stay because it appears before 5 in the original order, so the actual LIS length is 2 ([5,12] or [8,12]). Consequently 6‑2 = 4 moves are required. One optimal sequence: move 3 to front, move 5 after 3, move 7 after 5, move 10 after 8.
Constraints
- 1 <= arr.length <= 200000
- 0 <= arr[i] <= 10^9
- All arr[i] are distinct
Optimal Approach & Strategy
Compute the length of the longest increasing subsequence in O(n log n) and return n‑LIS as the minimal moves.
Brute Force Approach
Repeatedly scan the array, locate the smallest out‑of‑place element, and insert it at its correct earlier position, updating the array each time – O(n^2) time.
Code Solutions
function minShifts(arr){
const n = arr.length;
if(n===0) return 0;
const tails = [];
for(const x of arr){
let l=0, r=tails.length;
while(l<r){
const m = (l+r)>>1;
if(tails[m] < x) l=m+1; else r=m;
}
if(l===tails.length) tails.push_back ? tails.push(x) : tails.push(x);
else tails[l]=x;
}
return n - tails.length;
}
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(input.length){
const n = input[0];
const arr = input.slice(1,1+n);
console.log(minShifts(arr));
}#include <bits/stdc++.h>
using namespace std;
int minShifts(const vector<int>& arr) {
int n = arr.size();
if(n==0) return 0;
vector<int> tails;
tails.reserve(n);
for(int x: arr){
auto it = lower_bound(tails.begin(), tails.end(), x);
if(it==tails.end()) tails.push_back(x);
else *it = x;
}
return n - (int)tails.size();
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0; vector<int> a(n); for(int i=0;i<n;++i) cin>>a[i];
cout<<minShifts(a);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static int minShifts(int[] arr) {
int n = arr.length;
if(n==0) return 0;
int[] tails = new int[n];
int size = 0;
for(int x : arr){
int l = 0, r = size;
while(l < r){
int m = (l + r) >>> 1;
if(tails[m] < x) l = m + 1; else r = m;
}
if(l == size){
tails[size++] = x;
} else {
tails[l] = x;
}
}
return n - size;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
if(line == null || line.isEmpty()) return;
int n = Integer.parseInt(line.trim());
int[] arr = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) arr[i] = Integer.parseInt(st.nextToken());
System.out.print(minShifts(arr));
}
}def min_shifts(arr):
n = len(arr)
if n==0:
return 0
tails = []
for x in arr:
# binary search lower bound
lo, hi = 0, len(tails)
while lo < hi:
mid = (lo+hi)//2
if tails[mid] < x:
lo = mid+1
else:
hi = mid
if lo == len(tails):
tails.append(x)
else:
tails[lo] = x
return n - len(tails)
if __name__ == "__main__":
import sys
data = sys.stdin.read().strip().split()
if not data:
sys.exit()
n = int(data[0])
arr = list(map(int, data[1:1+n]))
print(min_shifts(arr))function minShifts(arr){
const n = arr.length;
if(n===0) return 0;
const tails = [];
for(const x of arr){
let l=0, r=tails.length;
while(l<r){
const m = (l+r)>>1;
if(tails[m] < x) l=m+1; else r=m;
}
if(l===tails.length) tails.push_back ? tails.push(x) : tails.push(x);
else tails[l]=x;
}
return n - tails.length;
}
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(input.length){
const n = input[0];
const arr = input.slice(1,1+n);
console.log(minShifts(arr));
}Asked in Top Tech Interviews
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