Summary Ranges - Practice Coding Problems

Summary Ranges - Problem

Array Easy

Summary Ranges - Transform a sorted array into compact range notation!

You're given a sorted unique integer array nums. Your mission is to convert this array into the smallest possible list of ranges that covers all numbers exactly once.

Range Format:
• "a->b" if the range spans multiple numbers (a ≠ b)
• "a" if the range contains only one number (a = b)

Goal: Return a list where each element from the original array appears in exactly one range, with no gaps or overlaps.

Example: [0,1,2,4,5,7] becomes ["0->2", "4->5", "7"]

Input & Output

example_1.py — Basic consecutive ranges

$ Input: [0,1,2,4,5,7]

› Output: ["0->2","4->5","7"]

💡 Note: The ranges are: [0,1,2] → "0->2", [4,5] → "4->5", and [7] → "7"

example_2.py — Mixed single elements and ranges

$ Input: [0,2,3,4,6,8,9]

› Output: ["0","2->4","6","8->9"]

💡 Note: The ranges are: [0] → "0", [2,3,4] → "2->4", [6] → "6", and [8,9] → "8->9"

example_3.py — Edge case empty array

$ Input: []

› Output: []

💡 Note: Empty input array results in empty output

Constraints

0 ≤ nums.length ≤ 20
-2³¹ ≤ nums[i] ≤ 2³¹ - 1
All the values of nums are unique
nums is sorted in ascending order

Visualization

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Understanding the Visualization

Start cataloging

Begin at the first book (number 0) and start a range

Find consecutive books

Books 0, 1, 2 are consecutive - they belong in one range entry

Detect gap

Book 4 is not consecutive to 2 (gap of 2) - close the current range

Create new range

Start a new range beginning with book 4, continue the process

Key Takeaway

🎯 Key Insight: Since the array is already sorted, we can detect range boundaries by simply checking if consecutive elements differ by exactly 1. When they don't, we know a range has ended!

Asked in

G Google 28 a Amazon 22 ⊞ Microsoft 15 f Meta 12

The optimal approach uses a single pass through the sorted array, tracking range boundaries by detecting gaps between consecutive elements. When nums[i] != nums[i-1] + 1, we know a range has ended and format it accordingly. This achieves O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass (Optimal)	O(n)	O(1)	Walk through the array once, tracking range boundaries
Brute Force (Nested Loops)	O(n²)	O(1)	For each starting position, find the longest consecutive sequence

Single Pass (Optimal) — Algorithm Steps

Initialize range start at first element
Iterate through array, checking if current element is consecutive to previous
When sequence breaks or we reach the end, format the current range
Add formatted range to result and start new range

Visualization

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Step-by-Step Walkthrough

Initialize

Set range start at index 0, begin iteration

Check consecutive

Compare each element with previous, continue if consecutive

Detect gap

When gap found (4 ≠ 2+1), close current range

Format range

Add "0->2" to result, start new range at current position

Code -

solution.c — C

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

char** summaryRanges(int* nums, int numsSize, int* returnSize) {
    *returnSize = 0;
    if (numsSize == 0) return NULL;
    
    char** result = (char**)malloc(numsSize * sizeof(char*));
    int start = 0;
    
    for (int i = 1; i <= numsSize; i++) {
        // Check if we've reached end or found a gap
        if (i == numsSize || nums[i] != nums[i - 1] + 1) {
            // Format and add current range
            result[*returnSize] = (char*)malloc(50 * sizeof(char));
            if (start == i - 1) {
                sprintf(result[*returnSize], "%d", nums[start]);
            } else {
                sprintf(result[*returnSize], "%d->%d", nums[start], nums[i - 1]);
            }
            
            (*returnSize)++;
            start = i;
        }
    }
    
    return result;
}

int main() {
    int nums[] = {0, 1, 2, 4, 5, 7};
    int returnSize;
    char** result = summaryRanges(nums, 6, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("\"%s\"", result[i]);
        if (i < returnSize - 1) printf(", ");
        free(result[i]);
    }
    printf("]\n");
    free(result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array, each element visited exactly once

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for tracking, output space not counted

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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