Minimum Number of Increasing Subsequence to Be Removed - Problem

Given an array of integers nums, you are allowed to perform the following operation any number of times:

Remove a strictly increasing subsequence from the array.

Your task is to find the minimum number of operations required to make the array empty.

A strictly increasing subsequence is a subsequence where each element is greater than the previous one.

Input & Output

Example 1 — Basic Case

$ Input: nums = [5,3,4,2,1]

› Output: 4

💡 Note: We need to partition the array into strictly increasing subsequences. We can remove [3,4] and then [5], [2], [1] separately. This requires 4 operations total.

Example 2 — Already Sorted

$ Input: nums = [1,2,3,4,5]

› Output: 1

💡 Note: The entire array is already strictly increasing, so we can remove it all in one operation.

Example 3 — All Same Elements

$ Input: nums = [5,5,5,5]

› Output: 4

💡 Note: Since we need strictly increasing subsequences, each element of value 5 must be removed separately, requiring 4 operations.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that this problem is equivalent to finding the minimum number of non-increasing subsequences needed to cover the array. The optimal approach uses binary search to efficiently maintain tails of decreasing subsequences. Time: O(n log n), Space: O(n)

Common Approaches

✓ Brute Force - Try All Combinations

⏱️ Time: O(2^n) Space: O(n)

This approach tries every possible way to divide the array into strictly increasing subsequences and finds the minimum number of such subsequences needed.

Greedy with Binary Search

⏱️ Time: O(n log n) Space: O(n)

The key insight is that the minimum number of operations equals the maximum number of elements that form a decreasing subsequence. We can use binary search to efficiently maintain multiple decreasing subsequences.

Brute Force - Try All Combinations — Algorithm Steps

Generate all possible subsequences
Check if each subsequence is strictly increasing
Find minimum partitioning into increasing subsequences

Visualization

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Step-by-Step Walkthrough

Generate Masks

Try all possible ways to select elements

Check Increasing

Verify if selected subsequence is strictly increasing

Recurse

Apply same process to remaining elements

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int isIncreasing(int* seq, int len) {
    for (int i = 1; i < len; i++) {
        if (seq[i] <= seq[i-1]) return 0;
    }
    return 1;
}

int minSubsequences(int* arr, int n) {
    if (n == 0) return 0;
    
    int minOps = INT_MAX;
    
    // Try all possible non-empty increasing subsequences
    for (int mask = 1; mask < (1 << n); mask++) {
        int subseq[20], remaining[20];
        int subLen = 0, remLen = 0;
        
        for (int i = 0; i < n; i++) {
            if (mask & (1 << i)) {
                subseq[subLen++] = arr[i];
            } else {
                remaining[remLen++] = arr[i];
            }
        }
        
        if (isIncreasing(subseq, subLen)) {
            int ops = 1 + minSubsequences(remaining, remLen);
            if (ops < minOps) minOps = ops;
        }
    }
    
    return minOps;
}

int solution(int* nums, int n) {
    return minSubsequences(nums, n);
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int n = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    printf("%d\n", solution(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Exponential time due to trying all possible combinations of subsequences

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack depth and temporary storage for subsequences

⚡ Linearithmic Space

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Minimum Number of Increasing Subsequence to Be Removed - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Combinations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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